Ch7_EigenValuesVector_Ex7.4.ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "id": "39765e0d",
   "metadata": {},
   "source": [
    "<b>Question 1</b> \n",
    "\n",
    "For the following matrices find an orthogonal matrix $Q$ which diagonalizes the given matrix. Also check that $Q^TAQ = D$ wjere $D$ is a diagonal matrix.\n",
    "\n",
    "(a)$A = \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "0&2\\\\\n",
    "\\end{pmatrix}\\quad$(b)$A = \\begin{pmatrix}\n",
    "1&1\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix}\\quad$(c)$A = \\begin{pmatrix}\n",
    "2&1\\\\\n",
    "1&2\\\\\n",
    "\\end{pmatrix}\\quad$(d)$A = \\begin{pmatrix}\n",
    "5&12\\\\\n",
    "12&-5\\\\\n",
    "\\end{pmatrix}$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "a8675f5e",
   "metadata": {},
   "source": [
    "<b> Solution (a)</b> \n",
    "\n",
    "Eigenvalue; $det(A - \\lambda I) = 0$<br>\n",
    "\n",
    "$\\Rightarrow det\\begin{pmatrix}\n",
    "1-\\lambda&0\\\\\n",
    "0&2-\\lambda\\\\\n",
    "\\end{pmatrix} = 0$\n",
    "\n",
    "$\\Rightarrow (1 - \\lambda)(2 - \\lambda) - 0 = 0$<br>\n",
    "$\\Rightarrow 2 - \\lambda - 2\\lambda + \\lambda^2 = 0 $<br>\n",
    "$\\Rightarrow \\lambda^2 - 3 \\lambda + 2 = 0$<br>\n",
    "$\\Rightarrow (\\lambda - 2) (\\lambda - 1) = 0$\n",
    "\n",
    "$\\Rightarrow \\lambda = 2, \\lambda = 1$\n",
    "\n",
    "For $\\lambda_{1} = 1$ eigenvector we have $(A - \\lambda I) u = 0$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "1-1&0\\\\\n",
    "0 & 2-1\\\\\n",
    "\\end{pmatrix}u = 0$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "0&0\\\\\n",
    "0&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow$ eigenvector $ u = \\begin{pmatrix}\n",
    "1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "For $\\lambda_{2} = 2$ eigenvector we have $(A - \\lambda I) u = 0$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "1-2&0\\\\\n",
    "0 & 2-2\\\\\n",
    "\\end{pmatrix}u = 0$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "-1&0\\\\\n",
    "0&0\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow$ eigenvector $ u = \\begin{pmatrix}\n",
    "0\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Diagonal matrix, $D$ is given by eigenvalues along the leading diagonal. Therefore, <br>\n",
    "$D = \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "0&2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Orthogonal matrix, Q, is given by $Q(u \\quad v)$ if normalized and orthogonal. Therefore, <br>\n",
    "$Q = \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "0&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Next, $Q^TAQ = \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "0&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "0&2\\\\\n",
    "\\end{pmatrix} \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "0&1\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "0&2\\\\\n",
    "\\end{pmatrix} = D$\n",
    "\n",
    "<b> Solution (b)</b> \n",
    "\n",
    "\n",
    "Here, eigevalues : $det(A - \\lambda I) =0$\n",
    "\n",
    "$\\Rightarrow det\\begin{pmatrix}\n",
    "1- \\lambda&1\\\\\n",
    "1 & 1 - \\lambda\\\\\n",
    "\\end{pmatrix} = 0$\n",
    "\n",
    "$\\Rightarrow (1 - \\lambda)(1 - \\lambda) - 1 = 0$<br>\n",
    "$\\Rightarrow (1 - \\lambda)^2 - 1 = 0$<br>\n",
    "$\\Rightarrow 1 + \\lambda^2 -2 \\lambda - 1 = 0$<br>\n",
    "$\\Rightarrow \\lambda(\\lambda -2) = 0$\n",
    "\n",
    "$\\Rightarrow \\lambda = 0$, $\\lambda - 2$\n",
    "\n",
    "For$\\lambda_{1} = 0$,\n",
    "eigenvector can be given by $(A - \\lambda I)u = 0$<br>\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "1-0&1\\\\\n",
    "1&1-0\\\\\n",
    "\\end{pmatrix}u = 0$<br>\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "1&1\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix}\\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$u = \\begin{pmatrix}\n",
    "1\\\\\n",
    "-1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "For eigenvector, when $\\lambda_{2} = 2$\n",
    "\n",
    "$\\begin{pmatrix}\n",
    "1-2&1\\\\\n",
    "1&1-2\\\\\n",
    "\\end{pmatrix} u = 0$<br>\n",
    "\n",
    "$\\begin{pmatrix}\n",
    "-1&1\\\\\n",
    "1&-1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$<br>\n",
    "\n",
    "$\\Rightarrow v = \\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$A$ is symmetric matrix, therefore, eigenvector $u$ and $v$ are orthogonal.\n",
    "\n",
    "Next, <br>\n",
    "\n",
    "$\\lVert u \\rVert^2 = \\lVert \\begin{pmatrix}\n",
    "1\\\\\n",
    "-1\\\\\n",
    "\\end{pmatrix}\\rVert^2 = \\begin{pmatrix}\n",
    "1\\\\\n",
    "-1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "1\\\\\n",
    "-1\\\\\n",
    "\\end{pmatrix} = 1^2 + (-1)^2 = 2$\n",
    "\n",
    "$\\Rightarrow \\lVert u \\rVert = \\sqrt{2}$\n",
    "\n",
    "$\\Rightarrow \\lVert v \\rVert^2 = \\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix} = 2$\n",
    "\n",
    "$\\Rightarrow \\lVert v \\rVert = \\sqrt{2}$\n",
    "\n",
    "$\\Rightarrow u = \\frac{1}{\\lVert u \\rVert}u = \\frac{1}{\\sqrt{2}}\\begin{pmatrix}\n",
    "1\\\\\n",
    "-1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow v = \\frac{1}{\\lVert v \\rVert} v = \\frac{1}{\\sqrt{2}}\\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Now, our orthogonal matrix, $Q$ is given by:\n",
    "$ Q = (\\widehat u \\widehat v) = \\frac{1}{\\sqrt{2}}\\begin{pmatrix}\n",
    "1&1\\\\\n",
    "-1&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$Q$ is an orthogonal matrix, therfore $Q^T = Q^{-1}$. \n",
    "\n",
    "$Q^TAQ = D$ or $QD =AQ$\n",
    "\n",
    "$AQ = \\frac{1}{\\sqrt{2}}\\begin{pmatrix}\n",
    "1&1\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix} \\begin{pmatrix}\n",
    "1&1\\\\\n",
    "-1&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{1}{\\sqrt{2}}\\begin{pmatrix}\n",
    "0&2\\\\\n",
    "0&2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "$QD = \\frac{1}{\\sqrt{2}}\\begin{pmatrix}\n",
    "0&0\\\\\n",
    "0&2\\\\\n",
    "\\end{pmatrix} \\begin{pmatrix}\n",
    "1&1\\\\\n",
    "-1&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{1}{\\sqrt{2}}\\begin{pmatrix}\n",
    "0&2\\\\\n",
    "0&2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "<b> Solution (c) </b>\n",
    "\n",
    "Eigenvalue $det(A - \\lambda I) = 0$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "2-\\lambda&1\\\\\n",
    "1&2-\\lambda\\\\\n",
    "\\end{pmatrix} = 0$\n",
    "\n",
    "$\\Rightarrow (2 - \\lambda)(2 - \\lambda) -1 = 0$<br>\n",
    "$\\Rightarrow 4 - 4\\lambda + \\lambda^2 -1 = 0$<br>\n",
    "$\\Rightarrow \\lambda^2 - 4\\lambda + 3 = 0$<br>\n",
    "$\\Rightarrow \\lambda - \\lambda -3\\lambda + 3 = 0$<br>\n",
    "$\\Rightarrow (\\lambda -1)(\\lambda -3) = 0$<br>\n",
    "$\\Rightarrow \\lambda_{1} =1$ and $\\lambda_{2} = 3$ \n",
    "\n",
    "For eigenvector, when $\\lambda_{1} = 1$<br>\n",
    "with $(A - \\lambda I)u = 0$<br>\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "2-1&1\\\\\n",
    "1&2-1\\\\\n",
    "\\end{pmatrix} \\cdot\\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "1&1\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix}$ \n",
    "\n",
    "$x + y = 0$<br>\n",
    "$\\Rightarrow x = 1, y = -1$\n",
    "\n",
    "Thus, eigenvector, $u = \\begin{pmatrix}\n",
    "1\\\\\n",
    "-1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "For eigenvector, when $\\lambda_{1} = 3$<br>\n",
    "with $(A - \\lambda I)u = 0$<br>\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "2-3&1\\\\\n",
    "1&2-3\\\\\n",
    "\\end{pmatrix} \\cdot\\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "-1&1\\\\\n",
    "1&-1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix}$ \n",
    "\n",
    "$-x + y = 0$<br>\n",
    "$x - y = 0$<br>\n",
    "\n",
    "Thus, eigenvector, $u = \\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$A$ is symmetric matrix, therefore, $u$ & $v$ are orthogonal.<br>\n",
    "\n",
    "Now, $\\lVert u \\rVert^2 = \\begin{pmatrix}\n",
    "1\\\\\n",
    "-1\\\\\n",
    "\\end{pmatrix}^2 = 1^2 + (-1)^2 = 2$\n",
    "\n",
    "$\\Rightarrow \\rVert u \\lVert = \\sqrt{2}$<br>\n",
    "and, $u = \\frac{1}{\\rVert u \\lVert}u = \\frac{1}{\\sqrt{2}}\\begin{pmatrix}\n",
    "1\\\\\n",
    "-1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Now, $\\lVert v \\rVert^2 = \\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}^2 = 1^2 + 1^2 = 2$\n",
    "\n",
    "$\\Rightarrow \\rVert v \\lVert = \\sqrt{2}$<br>\n",
    "and, $u = \\frac{1}{\\rVert v \\lVert}v = \\frac{1}{\\sqrt{2}}\\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix} $\n",
    "\n",
    "Next, our orthogonal matrix, $Q$ is given by $Q = (\\widehat u \\widehat v) = \\frac{1}{\\sqrt{2}}\\begin{pmatrix}\n",
    "1&1\\\\\n",
    "-1&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Diagonal matrix, $D$, is given by eigenvalues along the leading diagonal. Therefore, $D = \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "0&3\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$Q$ is an orthogonal matrix therefore, $Q^T = Q^{-1}$\n",
    "\n",
    "$\\Rightarrow Q^TAQ = D$<br>\n",
    "$\\Rightarrow QD = AQ$<br>\n",
    "\n",
    "Here, $QD = \\frac{1}{\\sqrt{2}}\\begin{pmatrix}\n",
    "1&1\\\\\n",
    "-1&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "1&0\\\\\n",
    "0&3\\\\\n",
    "\\end{pmatrix} = \\frac{1}{\\sqrt{2}}\\begin{pmatrix}\n",
    "1&3\\\\\n",
    "-1&3\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow AQ = \\frac{1}{\\sqrt{2}}\\begin{pmatrix}\n",
    "2&1\\\\\n",
    "1&2\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "1&1\\\\\n",
    "-1&1\\\\\n",
    "\\end{pmatrix} = \\frac{1}{\\sqrt{2}}\\begin{pmatrix}\n",
    "1&3\\\\\n",
    "-1&3\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "<b> Solution (d) </b>\n",
    "\n",
    "Eigenvalue $det(A - \\lambda) = 0$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "5 - \\lambda&12\\\\\n",
    "12& -5 - \\lambda\\\\\n",
    "\\end{pmatrix} = 0$<br>\n",
    "\n",
    "$\\Rightarrow (5 - \\lambda)(-5 - \\lambda) - 12(12) = 0$<br>\n",
    "$\\Rightarrow -25 -5 \\lambda + 5\\lambda - 144 = 0$<br>\n",
    "$\\Rightarrow \\lambda^2 - 169 = 0$<br>\n",
    "$\\Rightarrow \\lambda_{1} = 13$ and $\\lambda_{2} = -13$\n",
    "\n",
    "For eigenvector when $\\lambda_{1} = 13$\n",
    "\n",
    "$(A  - \\lambda I)u = 0$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "5 - 13&12\\\\\n",
    "12&-5-13\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "-8&12\\\\\n",
    "12&-18\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow -8x + 12y = 0$<br>\n",
    "and, $12x - 18y = 0$<br>\n",
    "\n",
    "$\\Rightarrow x = 3, y =2$\n",
    "\n",
    "$u = \\begin{pmatrix}\n",
    "3\\\\\n",
    "2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "For eigenvector when $\\lambda_{2} = -13$\n",
    "\n",
    "$(A  - \\lambda I)u = 0$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "5 -(-13)&12\\\\\n",
    "12&-5-(-13)\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "18&12\\\\\n",
    "12&8\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow 18x + 12y = 0$<br>\n",
    "and, $12x + 18y = 0$<br>\n",
    "\n",
    "$\\Rightarrow x = -2, y =3$\n",
    "\n",
    "$v = \\begin{pmatrix}\n",
    "-2\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Now, $A$ is symmetric, therefore, $u$ and $v$ are orthogonal. \n",
    "Next, <br>\n",
    "\n",
    "$\\rVert u \\lVert^2 = \\begin{pmatrix}\n",
    "3\\\\\n",
    "2\\\\\n",
    "\\end{pmatrix} = 3^2 + 2^2 = 13$\n",
    "\n",
    "$\\Rightarrow \\rVert u \\lVert = \\sqrt{13}$\n",
    "\n",
    "$\\Rightarrow u = \\frac{1}{\\rVert u \\lVert} u = \\frac{1}{\\sqrt{13}}\\begin{pmatrix}\n",
    "3\\\\\n",
    "2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "\n",
    "$\\rVert v \\lVert^2 = \\begin{pmatrix}\n",
    "-2\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix} = (-2)^2 + 3^2 = 13$\n",
    "\n",
    "$\\Rightarrow \\rVert v \\lVert = \\sqrt{13}$\n",
    "\n",
    "$\\Rightarrow v = \\frac{1}{\\rVert v \\lVert} v = \\frac{1}{\\sqrt{13}}\\begin{pmatrix}\n",
    "-2\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "With this, orthogonal matrix, $Q$, is given by: \n",
    "\n",
    "$Q = (\\widehat u \\widehat v) = \\frac{1}{\\sqrt{13}}\\begin{pmatrix}\n",
    "3&-2\\\\\n",
    "2&3\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Diagonal matrix, $D$ is given by eigenvalues along the leading diagonal. \n",
    "Then, <br>\n",
    "$D = \\begin{pmatrix}\n",
    "13&0\\\\\n",
    "0&-13\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$Q$ is an orthogonal matrix, therfore, $Q^T = A^{-1}$\n",
    "$\\Rightarrow Q^TAQ = D$ can be written as:<br>\n",
    "$QD = AQ$\n",
    "\n",
    "Now, $AQ = \\frac{1}{\\sqrt{13}}\\begin{pmatrix}\n",
    "5&12\\\\\n",
    "12&-5\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "3&-2\\\\\n",
    "2&3\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{1}{\\sqrt{13}}\\begin{pmatrix}\n",
    "15+24&-10+36\\\\\n",
    "36-10&-24-15\\\\\n",
    "\\end{pmatrix}$<br> $\\qquad= \\frac{1}{\\sqrt{13}}\\begin{pmatrix}\n",
    "39&26\\\\\n",
    "26&-39\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$QD = \\frac{1}{\\sqrt{13}}\\begin{pmatrix}\n",
    "3&-2\\\\\n",
    "2&3\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "13&0\\\\\n",
    "0&13\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{1}{\\sqrt{13}}\\begin{pmatrix}\n",
    "39&26\\\\\n",
    "26&-39\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Thus, $AQ = QD$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "c41d13cb",
   "metadata": {},
   "source": [
    "-----------------  "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "44ca42fe",
   "metadata": {},
   "source": [
    "<b> Question 2</b> \n",
    "\n",
    "For the following matrices find an orthogonal matrix $Q$ which diagonalizes the given matrix. Also check that, $Q^TAQ = D$ where $D$ is the diagonal matrix.\n",
    "\n",
    "(a)$A = \\begin{pmatrix}\n",
    "9&3\\\\\n",
    "3&1\\\\\n",
    "\\end{pmatrix}\\quad$(b)$A = \\begin{pmatrix}\n",
    "3&\\sqrt{2}\\\\\n",
    "\\sqrt{2}&2\\\\\n",
    "\\end{pmatrix}\\quad$(c)$A = \\begin{pmatrix}\n",
    "-5&\\sqrt{3}\\\\\n",
    "\\sqrt{3}&-3\\\\\n",
    "\\end{pmatrix}\\quad$(d)$A =  \\begin{pmatrix}\n",
    "4&\\sqrt{12}\\\\\n",
    "\\sqrt{12}&1\\\\\n",
    "\\end{pmatrix}\\quad$\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "077d71b3",
   "metadata": {},
   "source": [
    "<b> Solution (a)</b> \n",
    "\n",
    "Eigenvalue $det(A - \\lambda) = 0$\n",
    "\n",
    "$\\begin{pmatrix}\n",
    "9 - \\lambda & 3\\\\\n",
    "3 & 1 - \\lambda \\\\\n",
    "\\end{pmatrix} = 0$\n",
    "\n",
    "$\\Rightarrow (9 - \\lambda)(1 - \\lambda) - 9 = 0$<br>\n",
    "$\\Rightarrow 9 -9 \\lambda - \\lambda + \\lambda^2 - 9 = 0$<br>\n",
    "$\\Rightarrow \\lambda^2 - 10 \\lambda = 0$<br>\n",
    "$\\Rightarrow \\lambda(\\lambda - 10) = 0$<br>\n",
    "$\\Rightarrow \\lambda_{1} = 0$ or $\\lambda_{2} = 10$ \n",
    "\n",
    "For eigenvector, when $\\lambda_{1} = 0$\n",
    "\n",
    "$\\begin{pmatrix}\n",
    "9-0&3\\\\\n",
    "3&1-0\\\\\n",
    "\\end{pmatrix} \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "9&3\\\\\n",
    "3&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow 9x + 3y = 0$<br>\n",
    "and, $3x + 1y = 0$<br>\n",
    "\n",
    "$\\Rightarrow x = -1$ or $y = 3$\n",
    "\n",
    "$u = \\begin{pmatrix}\n",
    "-1\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "For eigenvector, when $\\lambda_{2} = 10$\n",
    "\n",
    "$\\begin{pmatrix}\n",
    "9-10&3\\\\\n",
    "3&1-10\\\\\n",
    "\\end{pmatrix} \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "-1&3\\\\\n",
    "3&-9\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow -x + 3y = 0$<br>\n",
    "and, $3x -9y = 0$<br>\n",
    "\n",
    "$\\Rightarrow x = 3$ or $y = 1$\n",
    "\n",
    "$v = \\begin{pmatrix}\n",
    "3\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "Now, $A$ is symmetric therefore $u$ and $v$ are orthogonal\n",
    "\n",
    "$\\rVert u \\lVert^2 = \\rVert \\begin{pmatrix}\n",
    "-1\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix} \\lVert^2 = (-1)^2 + 3^2 = 10$\n",
    "\n",
    "$\\rVert u \\lVert = \\sqrt{10}$<br>\n",
    "\n",
    "$\\Rightarrow u = \\frac{1}{\\rVert u \\lVert}u = \\frac{1}{\\sqrt{10}}\\begin{pmatrix}\n",
    "-1\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "$\\rVert v \\lVert^2 = \\rVert \\begin{pmatrix}\n",
    "3\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix} \\lVert^2 = 3^2 + 1^2 = 10$\n",
    "\n",
    "$\\rVert v \\lVert = \\sqrt{10}$<br>\n",
    "\n",
    "$\\Rightarrow v = \\frac{1}{\\rVert v \\lVert}v = \\frac{1}{\\sqrt{10}}\\begin{pmatrix}\n",
    "3\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "With this, orthogonal matrix, $Q$, is given by:\n",
    "\n",
    "$Q = (\\widehat u  \\widehat v) = \\frac{1}{\\sqrt{10}}\\begin{pmatrix}\n",
    "-1&3\\\\\n",
    "3&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Diagonal matrix, $D$, is given by eigenvalues along the leading diagonal. Then, <br>\n",
    "$D = \\begin{pmatrix}\n",
    "0&0\\\\\n",
    "0&10\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$Q$ is an orthogonal matrix therefore, $Q^T = Q^{-1}$\n",
    "\n",
    "$\\Rightarrow Q^TAQ = D$ can be written as: $QD = AQ$\n",
    "\n",
    "Now, $QD = \\frac{1}{\\sqrt{10}}\\begin{pmatrix}\n",
    "-1&3\\\\\n",
    "3&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "0&0\\\\\n",
    "0 &10\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{1}{\\sqrt{10}}\\begin{pmatrix}\n",
    "0&30\\\\\n",
    "0&10\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$AQ = \\frac{1}{\\sqrt{10}}\\begin{pmatrix}\n",
    "9&3\\\\\n",
    "3&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "-1&3\\\\\n",
    "3&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{1}{\\sqrt{10}}\\begin{pmatrix}\n",
    "-9+9&27+3\\\\\n",
    "-3+3&9+1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{1}{\\sqrt{10}}\\begin{pmatrix}\n",
    "0&30\\\\\n",
    "0&10\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "<b>Hence, $QD = AQ$</b>\n",
    "\n",
    "<b> Solution (b) </b>\n",
    "\n",
    "Eigenvalues $det(A - \\lambda) = 0$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "3 - \\lambda& \\sqrt{2}\\\\\n",
    "\\sqrt{2} & 2 - \\lambda\\\\\n",
    "\\end{pmatrix} = 0$\n",
    "\n",
    "$\\Rightarrow (3 - \\lambda)(2 - \\lambda) - \\sqrt{2}\\sqrt{2} = 0$<br>\n",
    "$\\Rightarrow 6 - 2\\lambda - 3\\lambda + \\lambda^2 - 2 = 0$<br>\n",
    "\n",
    "$\\Rightarrow \\lambda^2 - \\lambda -4 \\lambda + 4 = 0$<br>\n",
    "$\\Rightarrow \\lambda(\\lambda -1) - 4(\\lambda -1) = 0$<br>\n",
    "$\\Rightarrow (\\lambda -4)(\\lambda - 1) = 0$<br>\n",
    "$\\Rightarrow \\lambda = 4$ or $\\lambda = 1$\n",
    "\n",
    "For eigenvector, when $\\lambda_{1} = 4$\n",
    "\n",
    "$\\Rightarrow (A - \\lambda I)u = 0$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "3-4&\\sqrt{2}\\\\\n",
    "\\sqrt{2}&2-4\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "-1&\\sqrt{2}\\\\\n",
    "\\sqrt{2}&-2\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow -x + \\sqrt{2}y = 0$<br>\n",
    "and, $\\sqrt{2}x - 2y = 0$\n",
    "\n",
    "$\\Rightarrow x= 2$ or $y = \\sqrt{2}$\n",
    "\n",
    "$\\Rightarrow u = \\begin{pmatrix}\n",
    "2\\\\\n",
    "\\sqrt{2}\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "For eigenvector, when $\\lambda_{2} = 1$\n",
    "\n",
    "$\\Rightarrow (A - \\lambda I)u = 0$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "3-1&\\sqrt{2}\\\\\n",
    "\\sqrt{2}&2-1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "2&\\sqrt{2}\\\\\n",
    "\\sqrt{2}&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow -2x + \\sqrt{2}y = 0$<br>\n",
    "and, $\\sqrt{2}x + 1y = 0$\n",
    "\n",
    "$\\Rightarrow x= \\sqrt{2}$ or $y = -2$\n",
    "\n",
    "$\\Rightarrow v = \\begin{pmatrix}\n",
    "\\sqrt{2}\\\\\n",
    "-2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Now, $A$ is symmetric, therefore, $u$ and $v$ are orthogonal. <br>\n",
    "Next,<br>\n",
    "$\\lVert u \\rVert^2 = \\begin{pmatrix}\n",
    "2\\\\\n",
    "\\sqrt{2}\\\\\n",
    "\\end{pmatrix} = 2^2 + (\\sqrt{2})^2 = 6$<br>\n",
    "$\\Rightarrow \\lVert u \\rVert = \\sqrt{6}$<br>\n",
    "\n",
    "$u = \\frac{1}{\\lVert u \\rVert}u = \\frac{1}{\\sqrt{6}}\\begin{pmatrix}\n",
    "2\\\\\n",
    "\\sqrt{2}\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "$\\lVert v \\rVert^2 = \\begin{pmatrix}\n",
    "\\sqrt{2}\\\\\n",
    "-2\\\\\n",
    "\\end{pmatrix} = \\sqrt{2}^2 + (-2)^2 = 6$<br>\n",
    "$\\Rightarrow \\lVert v \\rVert = \\sqrt{6}$<br>\n",
    "\n",
    "$u = \\frac{1}{\\lVert v \\rVert}v = \\frac{1}{\\sqrt{6}}\\begin{pmatrix}\n",
    "\\sqrt{2}\\\\\n",
    "-2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "With this, orthogonal matrix, $Q$, is given by:<br>\n",
    "$Q = (\\widehat u \\widehat v) = \\frac{1}{\\sqrt{6}}\\begin{pmatrix}\n",
    "2&\\sqrt{2}\\\\\n",
    "\\sqrt{2}&-2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Diagonal matrix, $D$ is given eigenvalues along the leading diagonal\n",
    "\n",
    "$D = \\begin{pmatrix}\n",
    "4&0\\\\\n",
    "0&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$Q$ is an orthogonal matrix, therefore, $Q^T = Q^{-1}$\n",
    "\n",
    "$\\Rightarrow Q^TAQ = D$ can be written as: $QD = AQ$\n",
    "\n",
    "$QD = \\frac{1}{\\sqrt{6}}\\begin{pmatrix}\n",
    "2&\\sqrt{2}\\\\\n",
    "\\sqrt{2}&-2\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "4&0\\\\\n",
    "0&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{1}{\\sqrt{6}}\\begin{pmatrix}\n",
    "8&\\sqrt{2}\\\\\n",
    "4\\sqrt{2}&-2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "$AQ = \\frac{1}{\\sqrt{6}}\\begin{pmatrix}\n",
    "3&\\sqrt{2}\\\\\n",
    "\\sqrt{2}&2\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "2&\\sqrt{2}\\\\\n",
    "\\sqrt{2}&-2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{1}{\\sqrt{6}}\\begin{pmatrix}\n",
    "6+2 & 3\\sqrt{2} - 2\\sqrt{2}\\\\\n",
    "2\\sqrt{2} + 2\\sqrt{2} & 2 -4\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{1}{\\sqrt{6}}\\begin{pmatrix}\n",
    "8&\\sqrt{2}\\\\\n",
    "4\\sqrt{2}&-2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "<b> Solution (c) </b>\n",
    "\n",
    "Eigenvalue: $det(A - \\lambda) = 0$<br>\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "-5-\\lambda & \\sqrt{3}\\\\\n",
    "\\sqrt{3} & -3 - \\lambda\\\\\n",
    "\\end{pmatrix} = 0$\n",
    "\n",
    "$\\Rightarrow (-5 - \\lambda)(-3 - \\lambda) - \\sqrt{13}\\sqrt{13}$<br>\n",
    "\n",
    "$\\Rightarrow 15 + 5\\lambda + 3\\lambda + \\lambda^2 - 3= 0$<br>\n",
    "$\\Rightarrow \\lambda^2 + 8 \\lambda + 12 = 0$<br>\n",
    "$\\Rightarrow \\lambda^2 + 6 \\lambda + 2 \\lambda + 12 = 0$<br>\n",
    "$\\Rightarrow \\lambda(\\lambda + 6) + 2(\\lambda + 6) = 0$<br>\n",
    "$\\Rightarrow \\lambda = -6$ or $\\lambda = -2$\n",
    "\n",
    "For eigenvector when $\\lambda_{1} = -6$<br>\n",
    "with $(A - \\lambda I)u = 0$\n",
    "\n",
    "we get, $\\begin{pmatrix}\n",
    "-5+6& \\sqrt{3}\\\\\n",
    "\\sqrt{3}& -3+ 6\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "1&\\sqrt{3}\\\\\n",
    "\\sqrt{3}&-3\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow x + \\sqrt{3}y = 0$<br>\n",
    "and, $\\sqrt{3} + 3y = 0$\n",
    "\n",
    "$\\Rightarrow x = -3, y = \\sqrt{3}$\n",
    "\n",
    "$u = \\begin{pmatrix}\n",
    "-3\\\\\n",
    "\\sqrt{3}\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "For eigenvector when $\\lambda_{1} = -2$<br>\n",
    "with $(A - \\lambda I)u = 0$\n",
    "\n",
    "we get, $\\begin{pmatrix}\n",
    "-5+2& \\sqrt{3}\\\\\n",
    "\\sqrt{3}& -3+ 2\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "-3&\\sqrt{3}\\\\\n",
    "\\sqrt{3}&-1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow -3x + \\sqrt{3}y = 0$<br>\n",
    "and, $\\sqrt{3}x  -y = 0$\n",
    "\n",
    "$\\Rightarrow x = \\sqrt{3}, y = 3$\n",
    "\n",
    "$v = \\begin{pmatrix}\n",
    "\\sqrt{3}\\\\\n",
    "-3\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "Now, $A$ is symmetric, therefore, $u$ and $v$ are orthogonal. <br>\n",
    "Next,<br>\n",
    "$\\lVert u \\rVert^2 = \\begin{pmatrix}\n",
    "-3\\\\\n",
    "\\sqrt{3}\\\\\n",
    "\\end{pmatrix} = (-3)^2 + (\\sqrt{3})^2 = 12$<br>\n",
    "$\\Rightarrow \\lVert u \\rVert = \\sqrt{12}$<br>\n",
    "\n",
    "$u = \\frac{1}{\\lVert u \\rVert}u = \\frac{1}{\\sqrt{12}}\\begin{pmatrix}\n",
    "-3\\\\\n",
    "\\sqrt{3}\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "$\\lVert v \\rVert^2 = \\begin{pmatrix}\n",
    "\\sqrt{3}\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix} = \\sqrt{3}^2 + (3)^2 = 12$<br>\n",
    "$\\Rightarrow \\lVert v \\rVert = \\sqrt{12}$<br>\n",
    "\n",
    "$u = \\frac{1}{\\lVert v \\rVert}v = \\frac{1}{\\sqrt{12}}\\begin{pmatrix}\n",
    "\\sqrt{3}\\\\\n",
    "3\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "With this, orthogonal matrix, $Q$, is given by:<br>\n",
    "$Q = (\\widehat u \\widehat v) = \\frac{1}{\\sqrt{12}}\\begin{pmatrix}\n",
    "-3&\\sqrt{3}\\\\\n",
    "\\sqrt{3}&3\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Diagonal matrix, $D$ is given eigenvalues along the leading diagonal\n",
    "\n",
    "$D = \\begin{pmatrix}\n",
    "-6&0\\\\\n",
    "0&-2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$Q$ is an orthogonal matrix, therefore, $Q^T = Q^{-1}$\n",
    "\n",
    "$\\Rightarrow Q^TAQ = D$ can be written as: $QD = AQ$\n",
    "\n",
    "$QD = \\frac{1}{\\sqrt{12}}\\begin{pmatrix}\n",
    "-3&\\sqrt{13}\\\\\n",
    "\\sqrt{13}&3\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "-6&0\\\\\n",
    "0&-2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "$\\qquad = \\frac{1}{\\sqrt{12}}\\begin{pmatrix}\n",
    "18+0 & 0 - 2\\sqrt{3}\\\\\n",
    "-6\\sqrt{3}+0 & 0 -6\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{1}{12}\\begin{pmatrix}\n",
    "18&-2\\sqrt{3}\\\\\n",
    "-6\\sqrt{3}&-6\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$AQ = \\frac{1}{\\sqrt{12}}\\begin{pmatrix}\n",
    "-5&\\sqrt{13}\\\\\n",
    "\\sqrt{3}&-3\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "-3&\\sqrt{3}\\\\\n",
    "\\sqrt{3}&3\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "$\\qquad = \\frac{1}{\\sqrt{12}}\\begin{pmatrix}\n",
    "15 +3 & -5\\sqrt{3} + 3\\sqrt{3}\\\\\n",
    "-3\\sqrt{3} - 3\\sqrt{3}& 3 -9\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{1}{12}\\begin{pmatrix}\n",
    "18&-2\\sqrt{3}\\\\\n",
    "-6\\sqrt{3}&-6\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "<b>Solution (d) </b> \n",
    "\n",
    "Eigenvalue: $det(A - \\lambda I) = 0$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "5 - \\lambda & \\sqrt{12}\\\\\n",
    "\\sqrt{12} & 1 - \\lambda\\\\\n",
    "\\end{pmatrix} = 0$<br>\n",
    "\n",
    "$\\Rightarrow (5 - \\lambda)(1 - \\lambda) - \\sqrt{12}\\sqrt{12} = 0$<br>\n",
    "$\\Rightarrow 5- 5\\lambda - \\lambda + \\lambda^2 - 12 = 0$<br>\n",
    "$\\Rightarrow \\lambda^2 - 6\\lambda - 7 = 0$<br>\n",
    "$\\Rightarrow \\lambda^2 - 7 \\lambda + \\lambda - 7 =0 $<br>\n",
    "$\\Rightarrow (\\lambda + 1)(\\lambda - 7) = 0$\n",
    "\n",
    "$\\Rightarrow \\lambda = -1$ or $\\lambda = 7$\n",
    "\n",
    "For eigenvector when $\\lambda_{1} = -1$<br>\n",
    "with, $(A - \\lambda I)u = 0$<br>\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "5+1 & \\sqrt{12}\\\\\n",
    "\\sqrt{12} & 2 \\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "6 & \\sqrt{12}\\\\\n",
    "\\sqrt{12} & 2 \\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow 6x + \\sqrt{12}y = 0$<br>\n",
    "and, $\\sqrt{12}x + 2y = 0$\n",
    "\n",
    "$\\Rightarrow x = -2$ or $y = \\sqrt{12}$\n",
    "\n",
    "$u = \\begin{pmatrix}\n",
    "-2\\\\\n",
    "\\sqrt{12}\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "For eigenvector when $\\lambda_{2} = 7$<br>\n",
    "with, $(A - \\lambda I)u = 0$<br>\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "5-7 & \\sqrt{12}\\\\\n",
    "\\sqrt{12} & 1-7 \\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "-2 & \\sqrt{12}\\\\\n",
    "\\sqrt{12} & -6 \\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow -2x + \\sqrt{12}y = 0$<br>\n",
    "and, $\\sqrt{12}x + -6y = 0$\n",
    "\n",
    "$\\Rightarrow x = \\sqrt{12}$ or $y = 2$\n",
    "\n",
    "$v = \\begin{pmatrix}\n",
    "\\sqrt{12}\\\\\n",
    "2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Now $A$ is symmetric matrix, therefore, $u$ and $v$ are orthogonal\n",
    "\n",
    "$\\lVert u \\rVert^2 = \\begin{pmatrix}\n",
    "-2\\\\\n",
    "\\sqrt{12}\\\\\n",
    "\\end{pmatrix}^2 = (-2)^2 + \\sqrt{12}^2 = 16$<br>\n",
    "$\\Rightarrow  \\lVert u \\rVert = \\sqrt{16} = 4$<br>\n",
    "\n",
    "$\\Rightarrow u = \\frac{1}{\\lVert u \\rVert}u = \\frac{1}{4}\\begin{pmatrix}\n",
    "-2\\\\\n",
    "\\sqrt{12}\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "$\\lVert v \\rVert^2 = \\begin{pmatrix}\n",
    "\\sqrt{12}\\\\\n",
    "2\\\\\n",
    "\\end{pmatrix}^2 = (\\sqrt{12})^2 + 2^2 = 16$<br>\n",
    "$\\Rightarrow  \\lVert v \\rVert = \\sqrt{16} = 4$<br>\n",
    "\n",
    "$\\Rightarrow v = \\frac{1}{\\lVert v \\rVert}v = \\frac{1}{4}\\begin{pmatrix}\n",
    "\\sqrt{12}\\\\\n",
    "2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "With this, orthogonal matrix, $Q$ is given by:\n",
    "\n",
    "$Q = (\\widehat u \\widehat v) = \\frac{1}{4}\\begin{pmatrix}\n",
    "-2& \\sqrt{12}\\\\\n",
    "\\sqrt{12}& 2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{1}{2}\\begin{pmatrix}\n",
    "-1& \\sqrt{3}\\\\\n",
    "\\sqrt{3}& 1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Diagonal matrix, $D$ is given by eigenvalues along the leading diagonal. <br>\n",
    "Thus, $D = \\begin{pmatrix}\n",
    "-1&0\\\\\n",
    "0&7\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$Q$ is an orthogonal matrix, therefore $Q^T = Q^{-1}$\n",
    "\n",
    "$\\Rightarrow Q^TAQ = D$ can be written as $QD = AQ$\n",
    "\n",
    "Here, $QD =  \\frac{1}{2}\\begin{pmatrix}\n",
    "-1& \\sqrt{3}\\\\\n",
    "\\sqrt{3}& 1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "-1&0\\\\\n",
    "0&7\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{1}{2}\\begin{pmatrix}\n",
    "1&7\\sqrt{3}\\\\\n",
    "-\\sqrt{3} & 7\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$AQ =  \\frac{1}{2}\\begin{pmatrix}\n",
    "5& \\sqrt{12}\\\\\n",
    "\\sqrt{12}& 1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "-1&\\sqrt{3}\\\\\n",
    "\\sqrt{3}&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$AQ =  \\frac{1}{2}\\begin{pmatrix}\n",
    "5& 2\\sqrt{3}\\\\\n",
    "2\\sqrt{3}& 1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "-1&\\sqrt{3}\\\\\n",
    "\\sqrt{3}&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{1}{2}\\begin{pmatrix}\n",
    "1&7\\sqrt{3}\\\\\n",
    "-\\sqrt{3} & 7\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Hence, $QD = AQ$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "9dae3d5e",
   "metadata": {},
   "source": [
    "------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "ad49de05",
   "metadata": {},
   "source": [
    "<b>Question 3</b>\n",
    "\n",
    "For the following matrices find an orthogonal matrix $Q$ which diagonalizes the given matrix. BY using MATLAB or otherwise check that $Q^TAQ = D$ where $D$ is the diagonal matrix.\n",
    "\n",
    "(a)$A = \\begin{pmatrix}\n",
    "1&0&0\\\\\n",
    "0&2&0\\\\\n",
    "0&0&3\\\\\n",
    "\\end{pmatrix}\\quad$(b)$A = \\begin{pmatrix}\n",
    "2&2&2\\\\\n",
    "2&2&2\\\\\n",
    "2&2&2\\\\\n",
    "\\end{pmatrix}\\quad$(c)$A = \\begin{pmatrix}\n",
    "0&0&0\\\\\n",
    "0&1&1\\\\\n",
    "0&1&1\\\\\n",
    "\\end{pmatrix}$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "24e48aea",
   "metadata": {},
   "source": [
    "<b> Solution (a) </b> \n",
    "\n",
    "Since, our given matrix, $A = \\begin{pmatrix}\n",
    "1&0&0\\\\\n",
    "0&2&0\\\\\n",
    "0&0&3\\\\\n",
    "\\end{pmatrix}$ is also a diagonal matrix.\n",
    "\n",
    "Then, we need to find an orthogonal matrix, $Q$ such that<br> $Q^TAQ = A$\n",
    "\n",
    "If we take $Q$ equal to identity matrix, $I$, then <br>\n",
    "$Q^TAQ = I^TAI$<br>\n",
    "$\\qquad = A$\n",
    "\n",
    "Thus, we can say identity matrix, $I$ is the orthogonal matrix for $A$\n",
    "\n",
    "\n",
    "<b> Solution (b) </b>\n",
    "\n",
    "Eigenvalues: $det(A - \\lambda I) = 0$\n",
    "\n",
    "$\\Rightarrow (2 - \\lambda)\\begin{pmatrix}\n",
    "2-\\lambda&2\\\\\n",
    "2 & 2 - \\lambda\\\\\n",
    "\\end{pmatrix} - 2\\begin{pmatrix}\n",
    "2&2\\\\\n",
    "2&2-\\lambda\\\\\n",
    "\\end{pmatrix} + 2\\begin{pmatrix}\n",
    "2&2-\\lambda\\\\\n",
    "2&2\\\\\n",
    "\\end{pmatrix} = 0$\n",
    "\n",
    "$\\Rightarrow (2 - \\lambda)[(2- \\lambda)(2 - \\lambda) -4] -2[2(2 - \\lambda) -4] + 2[4 - 2(2-\\lambda)] = 0$<br>\n",
    "$\\Rightarrow (2 - \\lambda)(4 - 4\\lambda + \\lambda^2 + 4) - 2(4 - 2\\lambda - 4) + 2(4-4 + 2\\lambda) = 0$<br>\n",
    "$\\Rightarrow 2\\lambda^2 - 8 \\lambda - \\lambda^3 + 4\\lambda + 8 \\lambda = 0$<br>\n",
    "$\\Rightarrow \\lambda^3 - 6\\lambda^2 = 0$<br>\n",
    "$\\Rightarrow \\lambda^2 (\\lambda - 6) = 0$<br>\n",
    "$\\Rightarrow \\lambda^2 = 0$ and $\\lambda = 6$\n",
    "\n",
    "Hence, we can write eigenvalues as $\\lambda_{1} = 0, \\lambda_{2} = 0$ and $\\lambda_{3} = 6$\n",
    "\n",
    "For eigenvector when $\\lambda_{1} = 0$, <br>\n",
    "with $(A - \\lambda I)u = 0$<br>\n",
    "we get, <br>\n",
    "\n",
    "$\\begin{pmatrix}\n",
    "2-0&2&2\\\\\n",
    "2&2-0&2\\\\\n",
    "2&2&2-0\\\\\n",
    "\\end{pmatrix} \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow 2x + 2y + 2z  =0$<br>\n",
    "and, $2x + 2y + 2z  =0$<br>\n",
    "and, $2x + 2y + 2z  =0$\n",
    "\n",
    "This system of equations can have any values for $x, y$ and $z$. Howeverm here we are looking for orthogonal matrix, therefore we need to pick values of $x, y$ and $z$ for eigenvector $u$ and $v$ such that $u \\cdot v = 0$\n",
    "\n",
    "For example, if $u = \\begin{pmatrix}\n",
    "1\\\\\n",
    "-1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$, and $v = \\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "-2\\\\\n",
    "\\end{pmatrix}$ we have $u \\cdot v = 0$\n",
    "\n",
    "Next, for eigenvector when $\\lambda_{3} = 6$\n",
    "\n",
    "we get, $\\begin{pmatrix}\n",
    "2-6&2&2\\\\\n",
    "2&2-6&2\\\\\n",
    "2&2&2-6\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow -4x + 2y + 2z = 0$<br>\n",
    "and, $2x - 4y + 2z = 0$<br>\n",
    "and, $2x + 2y -4z = 0$<br>\n",
    "\n",
    "$\\Rightarrow x = y= z = 1$\n",
    "\n",
    "Therefore, third eigenvector $w = \\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Now, normalizing eigenvector\n",
    "\n",
    "$\\lVert u \\rVert^2 = \\lVert \\begin{pmatrix}\n",
    "1\\\\\n",
    "-1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}^2 \\rVert = 1^2 + (-1)^2 + 0^2 = 2$<br>\n",
    "\n",
    "$ \\lVert u \\rVert = \\sqrt{2}$<br>\n",
    "\n",
    "$\\Rightarrow \\widehat u = \\frac{1}{\\lVert u \\rVert} u = \\frac{1}{\\sqrt{2}}\\begin{pmatrix}\n",
    "1\\\\\n",
    "-1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "$\\lVert v \\rVert^2 = \\lVert \\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "-2\\\\\n",
    "\\end{pmatrix}^2 \\rVert = 1^2 + 1^2 + (-2)^2 = 6$<br>\n",
    "\n",
    "$ \\lVert v \\rVert = \\sqrt{6}$<br>\n",
    "\n",
    "$\\Rightarrow \\widehat v = \\frac{1}{\\lVert v \\rVert} v = \\frac{1}{\\sqrt{6}}\\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "-2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "$\\lVert w \\rVert^2 = \\lVert \\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}^2 \\rVert = 1^2 + 1^2 + 1^2 = 3$<br>\n",
    "\n",
    "$ \\lVert w \\rVert = \\sqrt{3}$<br>\n",
    "\n",
    "$\\Rightarrow \\widehat w = \\frac{1}{\\lVert w \\rVert} w = \\frac{1}{\\sqrt{3}}\\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Orthogonal matrix, $Q$ is given by $Q = (\\widehat u \\widehat v \\widehat w)$\n",
    "\n",
    "$\\Rightarrow Q = \\begin{pmatrix}\n",
    "1/\\sqrt{2}& 1/\\sqrt{6} & 1/\\sqrt{3}\\\\\n",
    "-1/\\sqrt{2}& 1/\\sqrt{6} & 1/\\sqrt{3}\\\\\n",
    "0& -2/\\sqrt{6} & 1/\\sqrt{3}\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "In this case, diagonal matrix, with eigenvalues along the leading diagonal would be :\n",
    "\n",
    "$D = \\begin{pmatrix}\n",
    "0&0&0\\\\\n",
    "0&0&0\\\\\n",
    "0&0&6\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Next using Scipy to get $Q^TAQ$ and check if it's equal to $D$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "id": "738c22c7",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Resultant matrix of Q^TAQ\n",
      "[[ 0.  0.  0.]\n",
      " [ 0.  0.  0.]\n",
      " [-0.  0.  6.]]\n"
     ]
    }
   ],
   "source": [
    "import numpy as np\n",
    "\n",
    "matrix1 = np.array([\n",
    "    [1/np.sqrt(2), 1/np.sqrt(6), 1/np.sqrt(3)],\n",
    "    [-1/np.sqrt(2), 1/np.sqrt(6), 1/np.sqrt(3)],\n",
    "    [0, -2/np.sqrt(6), 1/np.sqrt(3)]\n",
    "])\n",
    "\n",
    "matrix2 = np.array([\n",
    "    [2, 2, 2],\n",
    "    [2, 2, 2],\n",
    "    [2, 2, 2]\n",
    "])\n",
    "\n",
    "matrix3 = np.array([\n",
    "    [1/np.sqrt(2), 1/np.sqrt(6), 1/np.sqrt(3)],\n",
    "    [-1/np.sqrt(2), 1/np.sqrt(6), 1/np.sqrt(3)],\n",
    "    [0, -2/np.sqrt(6), 1/np.sqrt(3)]\n",
    "])\n",
    "\n",
    "result = np.matmul(np.matmul(matrix1.T, matrix2), matrix3)\n",
    "\n",
    "# Round the elements to two decimal points\n",
    "result_rounded = np.round(result, 2)\n",
    "\n",
    "print(\"Resultant matrix of Q^TAQ\")\n",
    "print(result_rounded)\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "7c55c7f5",
   "metadata": {},
   "source": [
    "<b> Solution (c)</b>\n",
    "\n",
    "Eigenvalues: $det(A - \\lambda I) = 0$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "-\\lambda &0 &0\\\\\n",
    "0 & 1-\\lambda&1\\\\\n",
    "0&1&1-\\lambda\\\\\n",
    "\\end{pmatrix} = 0$<br>\n",
    "\n",
    "$\\Rightarrow -\\lambda[(1-\\lambda)(1-\\lambda) -1] - 0[0(1-\\lambda) - 0(1)] + 0[0(1) -0(1)] = 0$<br>\n",
    "\n",
    "$\\Rightarrow -\\lambda(1 - 2\\lambda + \\lambda^2 -1) = 0$<br>\n",
    "$\\Rightarrow -\\lambda^3 + 2\\lambda^2 = 0$<br>\n",
    "$\\Rightarrow \\lambda^3 - 2\\lambda^2 = 0$<br>\n",
    "$\\Rightarrow \\lambda^2 (\\lambda -2) = 0$<br>\n",
    "\n",
    "$\\Rightarrow \\lambda^2 = 0$ or $\\lambda = 2$\n",
    "\n",
    "Here, we can write eigenvalues as $\\lambda_{1} = 0, \\lambda_{2} = 0$ and $\\lambda_{3} =2$\n",
    "\n",
    "For eigenvectors when $\\lambda_{1} = 0$<br>\n",
    "with $(A - \\lambda I) u = 0$, \n",
    "\n",
    "we get, <br>\n",
    "$\\begin{pmatrix}\n",
    "0-0&0&0\\\\\n",
    "0&1-0&1\\\\\n",
    "0&1&1-0\\\\\n",
    "\\end{pmatrix} \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow 0x + y + z = 0$<br>\n",
    "and, $0x + y + z = 0$\n",
    "\n",
    "$\\Rightarrow y = -z$ or $u = \\begin{pmatrix}\n",
    "0\\\\\n",
    "-1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "However, we are looking for orthogonal matrix, we need to pick values of $x, y$ and $z$ for $v$ such that $u \\cdot v = 0$ if $v = \\begin{pmatrix}\n",
    "1\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$ then we have $u \\cdot v = 0$\n",
    "\n",
    "Next, for eigenvector when $\\lambda_{3} = 2$, \n",
    "\n",
    "we get $\\begin{pmatrix}\n",
    "0-2&0&0\\\\\n",
    "0&1-2&1\\\\\n",
    "0&1&1-2\\\\\n",
    "\\end{pmatrix} \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow -2x = 0$<br>\n",
    "and, $-y + z = 0$<br>\n",
    "and, $y - z = 0$\n",
    "\n",
    "$\\Rightarrow x  =0$ and $y =z$ where $y$ and $z$ can be any non-zero\n",
    "\n",
    "Therefore, $w = \\begin{pmatrix}\n",
    "0\\\\\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Now, normalizing eigenvectors\n",
    "\n",
    "$\\lVert u \\rVert^2 = \\lVert \\begin{pmatrix}\n",
    "0\\\\\n",
    "-1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix} \\rVert^2 = 0^2 + (-1)^2 + 1^2 = 2$\n",
    "\n",
    "$\\Rightarrow \\lVert u \\rVert = \\sqrt{2}$\n",
    "\n",
    "$\\Rightarrow \\widehat u = \\frac{1}{\\lVert u \\rVert}u = \\frac{1}{\\sqrt{2}}  \\begin{pmatrix}\n",
    "0\\\\\n",
    "-1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\lVert v \\rVert^2 = \\lVert \\begin{pmatrix}\n",
    "1\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix} \\rVert^2 = 1^2 + 0^2 + 0^2 = 1$\n",
    "\n",
    "$\\Rightarrow \\lVert v \\rVert = \\sqrt{1} = 1$\n",
    "\n",
    "$\\Rightarrow \\widehat v = \\frac{1}{\\lVert v \\rVert}v = \\begin{pmatrix}\n",
    "1\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\lVert w \\rVert^2 = \\lVert \\begin{pmatrix}\n",
    "0\\\\\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix} \\rVert^2 = 0^2 + 1^2 + 1^2 = 2$\n",
    "\n",
    "$\\Rightarrow \\lVert w \\rVert = \\sqrt{2}$\n",
    "\n",
    "$\\Rightarrow \\widehat w = \\frac{1}{\\lVert w \\rVert}w = \\frac{1}{\\sqrt{2}}  \\begin{pmatrix}\n",
    "0\\\\\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Orthogonal matrix, $Q$ is given by $Q = (\\widehat u \\widehat v \\widehat w)$\n",
    "\n",
    "Therefore, $Q = \\begin{pmatrix}\n",
    "0&1&0\\\\\n",
    "-1/\\sqrt{2}&0&1/\\sqrt{2}\\\\\n",
    "1/\\sqrt{2}&0&1/\\sqrt{2}\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "In this case, diagonal matrix, with eigenvalues along the leading diagonal would be :\n",
    "\n",
    "$D = \\begin{pmatrix}\n",
    "0&0&0\\\\\n",
    "0&0&0\\\\\n",
    "0&0&2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Next using Scipy to get $Q^TAQ$ and check if it's equal to $D$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "id": "32d13699",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Resultant matrix:\n",
      "[[0. 0. 0.]\n",
      " [0. 0. 0.]\n",
      " [0. 0. 2.]]\n"
     ]
    }
   ],
   "source": [
    "import numpy as np\n",
    "\n",
    "# Define the matrices\n",
    "matrix1 = np.array([\n",
    "    [0, 1, 0],\n",
    "    [-1/np.sqrt(2), 0, 1/np.sqrt(2)],\n",
    "    [1/np.sqrt(2), 0, 1/np.sqrt(2)]\n",
    "])\n",
    "\n",
    "matrix2 = np.array([\n",
    "    [0, 0, 0],\n",
    "    [0, 1, 1],\n",
    "    [0, 1, 1]\n",
    "])\n",
    "\n",
    "matrix3 = np.array([\n",
    "    [0, 1, 0],\n",
    "    [-1/np.sqrt(2), 0, 1/np.sqrt(2)],\n",
    "    [1/np.sqrt(2), 0, 1/np.sqrt(2)]\n",
    "])\n",
    "\n",
    "# Perform the matrix multiplications\n",
    "result = np.matmul(np.matmul(matrix1.T, matrix2), matrix3)\n",
    "result_rounded = np.round(result, 2)\n",
    "\n",
    "print(\"Resultant matrix:\")\n",
    "print(result_rounded)\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "3a76b1cd",
   "metadata": {},
   "source": [
    "------------------------------- "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "3bfbac17",
   "metadata": {},
   "source": [
    "<b> Question 4 </b> \n",
    "\n",
    "For the following matrices find an orthogonal matrix $Q$ which diagonalizes the given matrix. BY using MATLAB or otherwise check that $Q^TAQ = D$.\n",
    "\n",
    "(a)$A = \\begin{pmatrix}\n",
    "1&2&2\\\\\n",
    "2&1&2\\\\\n",
    "2&2&1\\\\\n",
    "\\end{pmatrix}\\quad$(b)$A = \\begin{pmatrix}\n",
    "2&1&1\\\\\n",
    "1&2&1\\\\\n",
    "1&1&2\\\\\n",
    "\\end{pmatrix}\\quad$(c)$A = \\begin{pmatrix}\n",
    "-5&4&2\\\\\n",
    "4&-5&2\\\\\n",
    "2&2&-8\\\\\n",
    "\\end{pmatrix}$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "f09189d2",
   "metadata": {},
   "source": [
    "<b>Solution (a) </b>\n",
    "\n",
    "Eigenvalue: $det(A - \\lambda I) = 0$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "1-\\lambda& 2 & 2\\\\\n",
    "2 & 1 - \\lambda & 2\\\\\n",
    "2 & 2 & 1 - \\lambda\\\\\n",
    "\\end{pmatrix} = 0$\n",
    "\n",
    "$\\Rightarrow (1- \\lambda)[(1 - \\lambda)(1 - \\lambda) - 2(2)] - 2[2(1- \\lambda) -2(2)] + 2[2(2) - 2(1 - \\lambda)] = 0$<br>\n",
    "\n",
    "$\\Rightarrow (1- \\lambda)[(1 - \\lambda)^2 - 4] -2[2 - 2\\lambda -4] + 2[4 -2 + 2\\lambda] = 0$<br>\n",
    "\n",
    "$\\Rightarrow (1- \\lambda)(1 + \\lambda^2 -2\\lambda - 4) -2(-2\\lambda -2) + 2(2\\lambda + 2) = 0$<br>\n",
    "\n",
    "$\\Rightarrow (1- \\lambda)(\\lambda^2 - 2\\lambda -3) + 4\\lambda + 4 +4\\lambda + 4 = 0$<br>\n",
    "\n",
    "$\\Rightarrow \\lambda^2 - 2\\lambda -3 - \\lambda^3 + 2\\lambda^2 + 3\\lambda + 8\\lambda +8 = 0$<br>\n",
    "\n",
    "$\\Rightarrow -\\lambda^3 + 3\\lambda^2 + 9\\lambda + 5 = 0$<br>\n",
    "$\\Rightarrow \\lambda^3 - 3\\lambda^2 - 9\\lambda - 5 = 0$<br>\n",
    "\n",
    "$\\Rightarrow \\lambda^3 -5\\lambda^2 + 2\\lambda^2 + \\lambda - 10\\lambda -5 = 0$\n",
    "\n",
    "$\\Rightarrow \\lambda^3 + 2\\lambda^2 + \\lambda -5\\lambda^2 - 10\\lambda -5 = 0$\n",
    "\n",
    "$\\Rightarrow \\lambda(\\lambda^2 + 2\\lambda + 1) - 5(\\lambda^2 + 2\\lambda + 1) = 0$\n",
    "\n",
    "$\\Rightarrow (\\lambda - 5)(\\lambda^2 + 2\\lambda + 1) =0 $\n",
    "\n",
    "$\\Rightarrow (\\lambda - 5)(\\lambda + 1)^2 =0 $\n",
    "\n",
    "$\\Rightarrow \\lambda = 5$ and $\\lambda = -1$ with multiplicity of 2.\n",
    "\n",
    "<b>Hence, our eigenvalues are $\\lambda_{1} = 5, \\lambda_{2} = -1$ and $\\lambda_{3} = -1$</b>\n",
    "\n",
    "Now for eigenvector when $\\lambda = 5$<br>\n",
    "with $(A - \\lambda I) u = 0$<br>\n",
    "we get, $\\begin{pmatrix}\n",
    "1-5&2&2\\\\\n",
    "2&1-5&2\\\\\n",
    "2&2&1-5\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "-4&2&2\\\\\n",
    "2&-4&2\\\\\n",
    "2&2&-4\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow -4x + 2y + 2z = 0$<br>\n",
    "and, $2x - 4y + 2z = 0$<br>\n",
    "and, $2x + 2y -4z = 0$\n",
    "\n",
    "$\\Rightarrow x = y = z = 1$. Thus eigenvector $u = \\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Next, for eigenvalues when $\\lambda = -1$<br>\n",
    "we get, $\\begin{pmatrix}\n",
    "1-(-1)&2&2\\\\\n",
    "2&1-(-1)&2\\\\\n",
    "2&2&1-(-1)\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow \\begin{pmatrix}\n",
    "2&2&2\\\\\n",
    "2&2&2\\\\\n",
    "2&2&2\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "z\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow 2x + 2y + 2z  =0$<br>\n",
    "and, $2x + 2y + 2z  =0$<br>\n",
    "and, $2x + 2y + 2z  =0$\n",
    "\n",
    "This system of equations can have any values for $x, y$ and $z$. Howeverm here we are looking for orthogonal matrix, therefore we need to pick values of $x, y$ and $z$ for eigenvector $v$ and $w$ such that $v \\cdot w = 0$\n",
    "\n",
    "For example, if $v = \\begin{pmatrix}\n",
    "1\\\\\n",
    "-1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$, and $w = \\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "-2\\\\\n",
    "\\end{pmatrix}$ we have $v \\cdot w = 0$\n",
    "\n",
    "Now, normalizing eigenvectors, \n",
    "\n",
    "$\\lVert u \\rVert^2 = \\lVert \\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix} \\rVert^2 = 1^2 +1^2 +1^2 = 3$\n",
    "\n",
    "$\\Rightarrow \\lVert u \\rVert = \\sqrt{3}$<br>\n",
    "\n",
    "$\\Rightarrow \\widehat u = \\frac{1}{\\lVert u \\rVert} u = \\frac{1}{\\sqrt{3}}\\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "$\\lVert v \\rVert^2 = \\lVert \\begin{pmatrix}\n",
    "1\\\\\n",
    "-1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix} \\rVert^2 = 1^2 + (-1)^2 +0^2 = 2$\n",
    "\n",
    "$\\Rightarrow \\lVert v \\rVert = \\sqrt{2}$<br>\n",
    "\n",
    "$\\Rightarrow \\widehat v = \\frac{1}{\\lVert v \\rVert} v = \\frac{1}{\\sqrt{2}}\\begin{pmatrix}\n",
    "1\\\\\n",
    "-1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "$\\lVert w \\rVert^2 = \\lVert \\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "-2\\\\\n",
    "\\end{pmatrix} \\rVert^2 = 1^2 + 1^2 +2^2 = 6$\n",
    "\n",
    "$\\Rightarrow \\lVert w \\rVert = \\sqrt{6}$<br>\n",
    "\n",
    "$\\Rightarrow \\widehat w = \\frac{1}{\\lVert w \\rVert} w = \\frac{1}{\\sqrt{6}}\\begin{pmatrix}\n",
    "1\\\\\n",
    "1\\\\\n",
    "-2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Orthogonal matrix, $Q$ is given by $Q = (\\widehat u \\widehat v \\widehat w)$\n",
    "\n",
    "Thus, we can write $Q = \\begin{pmatrix}\n",
    "1/\\sqrt{3}&1/\\sqrt{2}&1/\\sqrt{6}\\\\\n",
    "1/\\sqrt{3}&-1/\\sqrt{2}&1/\\sqrt{6}\\\\\n",
    "1/\\sqrt{3}&0&-2/\\sqrt{6}\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "In this case, diagonal matrix, with eigenvalues along the leading diagonal would be :\n",
    "\n",
    "$D = \\begin{pmatrix}\n",
    "5&0&0\\\\\n",
    "0&-1&0\\\\\n",
    "0&0&-1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Next using Scipy to get $Q^TAQ$ and check if it's equal to $D$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "id": "0ca99ea9",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Resultant matrix:\n",
      "[[ 5.  0. -0.]\n",
      " [ 0. -1. -0.]\n",
      " [-0.  0. -1.]]\n"
     ]
    }
   ],
   "source": [
    "matrix1 = np.array([\n",
    "    [1/np.sqrt(3), 1/np.sqrt(2), 1/np.sqrt(6)],\n",
    "    [1/np.sqrt(3), -1/np.sqrt(2), 1/np.sqrt(6)],\n",
    "    [1/np.sqrt(3), 0, -2/np.sqrt(6)]\n",
    "])\n",
    "\n",
    "matrix2 = np.array([\n",
    "    [1, 2, 2],\n",
    "    [2, 1, 2],\n",
    "    [2, 2, 1]\n",
    "])\n",
    "\n",
    "matrix3 = np.array([\n",
    "    [1/np.sqrt(3), 1/np.sqrt(2), 1/np.sqrt(6)],\n",
    "    [1/np.sqrt(3), -1/np.sqrt(2), 1/np.sqrt(6)],\n",
    "    [1/np.sqrt(3), 0, -2/np.sqrt(6)]\n",
    "])\n",
    "\n",
    "# Perform the matrix multiplications\n",
    "result = np.matmul(np.matmul(matrix1.T, matrix2), matrix3)\n",
    "result_rounded = np.round(result, 2)\n",
    "\n",
    "print(\"Resultant matrix:\")\n",
    "print(result_rounded)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "bb32a868",
   "metadata": {},
   "source": [
    "------------------------ "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "e5192258",
   "metadata": {},
   "source": [
    "<b> Question 5</b> \n",
    "\n",
    "Let $A = \\begin{pmatrix}\n",
    "1&1\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix}$. Show that $A^{10} = 2^9A$. Also prove that $A^m = 2^{m-1}A$ where $m$ is a positive integer."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "c409d73a",
   "metadata": {},
   "source": [
    "<b> Solution </b>\n",
    "\n",
    "In Question <b>1(b)</b> we already have calculated  orthogonal matrix, $Q$ and diagonal matrix, $Q$ for this matrix $A$.\n",
    "We have, <br>\n",
    "\n",
    "$Q = \\frac{1}{\\sqrt{2}}\\begin{pmatrix}\n",
    "1&1\\\\\n",
    "-1&1\\\\\n",
    "\\end{pmatrix}$, and $D = \\begin{pmatrix}\n",
    "0&0\\\\\n",
    "0&2\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "We also know that, <br>\n",
    "\n",
    "$A^m = QD^mQ^T$\n",
    "\n",
    "Therefore, <br>\n",
    "\n",
    "$A^{10} = \\frac{1}{\\sqrt{2}}\\begin{pmatrix}\n",
    "1&1\\\\\n",
    "-1&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "0&0\\\\\n",
    "0&2\\\\\n",
    "\\end{pmatrix}^{10} \\cdot \\frac{1}{\\sqrt{2}}\\begin{pmatrix}\n",
    "1&1\\\\\n",
    "-1&1\\\\\n",
    "\\end{pmatrix}^T$\n",
    "\n",
    "$\\qquad = \\frac{1}{\\sqrt{2}}\\frac{1}{\\sqrt{2}}\\begin{pmatrix}\n",
    "1&1\\\\\n",
    "-1&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "0&0\\\\\n",
    "0&2^{10}\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "1&-1\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{1}{2}\\begin{pmatrix}\n",
    "1&1\\\\\n",
    "-1&1\\\\\n",
    "\\end{pmatrix} \\cdot 2^{10} \\begin{pmatrix}\n",
    "0&0\\\\\n",
    "0&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "1&-1\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{2^{10}}{2}\\begin{pmatrix}\n",
    "1&1\\\\\n",
    "-1&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "0&0\\\\\n",
    "0&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "1&-1\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = 2^9 \\begin{pmatrix}\n",
    "0&1\\\\\n",
    "0&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "1&-1\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = 2^9  \\begin{pmatrix}\n",
    "1&1\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = 2^9A$\n",
    "\n",
    "<b>Hence, proved $A^{10} = 2^9 A$</b>\n",
    "\n",
    "Next, <br>\n",
    "\n",
    "$A^m = QD^mQ^T $<br>\n",
    "$\\qquad = \\frac{1}{\\sqrt{2}}\\begin{pmatrix}\n",
    "1&1\\\\\n",
    "-1&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "0&0\\\\\n",
    "0&2\\\\\n",
    "\\end{pmatrix}^m \\cdot \\frac{1}{\\sqrt{2}}\\begin{pmatrix}\n",
    "1&1\\\\\n",
    "-1&1\\\\\n",
    "\\end{pmatrix}^T$\n",
    "\n",
    "\n",
    "$\\qquad = \\frac{1}{\\sqrt{2}}\\frac{1}{\\sqrt{2}}\\begin{pmatrix}\n",
    "1&1\\\\\n",
    "-1&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "0&0\\\\\n",
    "0&2^m\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "1&-1\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{1}{2}\\begin{pmatrix}\n",
    "1&1\\\\\n",
    "-1&1\\\\\n",
    "\\end{pmatrix} \\cdot 2^m \\begin{pmatrix}\n",
    "0&0\\\\\n",
    "0&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "1&-1\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = \\frac{2^m}{2}\\begin{pmatrix}\n",
    "1&1\\\\\n",
    "-1&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "0&0\\\\\n",
    "0&1\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "1&-1\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "$\\qquad = 2^{m-1}\\begin{pmatrix}\n",
    "1&1\\\\\n",
    "1&1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\qquad = 2^{m-1}A$\n",
    "\n",
    "<b>Hence, proved $A^m = 2^{m-1}$</b>"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "c791e043",
   "metadata": {},
   "source": [
    "----------------------------- "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "226099d5",
   "metadata": {},
   "source": [
    "<b> Question 6</b> \n",
    "\n",
    "Show that if $A$ is a diagonal matrix then orthogonal diagonalizing matrix $Q = 1$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "96acf053",
   "metadata": {},
   "source": [
    "<b> Solution </b>\n",
    "\n",
    "We have $A$ as diagonal matrix.<br>\n",
    "Let's take $Q = I$<br>\n",
    "Then, <br>\n",
    "$Q^{-1}AQ = I^{-1} AI$<br>\n",
    "$\\qquad = IAI$<br>\n",
    "$\\qquad = A$\n",
    "\n",
    "$\\Rightarrow I$ orthogonally diagonalizes $A$.\n",
    "\n",
    "Hence, proved if $A$ is a diagonal matrix, then orthogonally diagonalizing matrix, $Q$ is equal to identity matrix, $I$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "3a84737a",
   "metadata": {},
   "source": [
    "----------------------------- "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "e8385be6",
   "metadata": {},
   "source": [
    "<b> Question 7</b> \n",
    "\n",
    "Prove that (a) the zero matrix <b>O</b> and (b) the identity matrix $I$ are orthogonally diagonalizable"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "1ddfd02b",
   "metadata": {},
   "source": [
    "<b> Solution (a)</b>\n",
    "\n",
    "Lets suppose our zero matrix, <b>O</b> is $3 \\times 3$.\n",
    "\n",
    "Then, <br>\n",
    "Eigenvalues are $\\lambda_{1} = \\lambda_{2} = \\lambda_{3} = 0$\n",
    "\n",
    "Implying, eigenvectors as:\n",
    "\n",
    "$u = \\begin{pmatrix}\n",
    "1\\\\\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}, v = \\begin{pmatrix}\n",
    "0\\\\\n",
    "1\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$, and $w = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "1\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Normalized values of $u, v, w$ would be the same.\n",
    "\n",
    "Therefore now, orthogonal matrix, $Q$ can be given by :\n",
    "\n",
    "$Q = (\\widehat u \\widehat v \\widehat w )$\n",
    "\n",
    "$\\Rightarrow Q = \\begin{pmatrix}\n",
    "1&0&0\\\\\n",
    "0&1&0\\\\\n",
    "0&0&1\\\\\n",
    "\\end{pmatrix} = I$\n",
    "\n",
    "Then, $Q^{-1}OQ = I^{1}OI = IOI = O$\n",
    "\n",
    "Hence, orthogonally diagonalizable.\n",
    "\n",
    "<b> Solution (b)</b>\n",
    "\n",
    "Since, any identity matrix, $I$ is also a diagonal matrix.\n",
    "Therefore, from previous solution of question 6, we can state identity matrix is also a diagonalizable matrix, and the orthogonal matrix, $Q = I$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "fd898e55",
   "metadata": {},
   "source": [
    "------------------------- "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "3180eda6",
   "metadata": {},
   "source": [
    "<b> Question 8</b>\n",
    "\n",
    "Prove that $A = \\begin{pmatrix}\n",
    "a&b\\\\\n",
    "b&c\\\\\n",
    "\\end{pmatrix} \\ne o$ is orthogonally diagonalizable and find the orthogonal matrix $Q$ which diagonalizes the matrix $A$."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "731f56fb",
   "metadata": {},
   "source": [
    "<b> Solution </b>\n",
    "\n",
    "Here, $A = \\begin{pmatrix}\n",
    "a&b\\\\\n",
    "b&c\\\\\n",
    "\\end{pmatrix} = A^T$\n",
    "\n",
    "Therefore, $A$ is a symmetric matrix, thus orthogonally diagonalizable\n",
    "\n",
    "Next, Eigenvalues: $det(A - \\lambda) = 0$\n",
    "\n",
    "$det\\begin{pmatrix}\n",
    "a - \\lambda & b\\\\\n",
    "b & c - \\lambda\\\\\n",
    "\\end{pmatrix} = 0$\n",
    "\n",
    "$\\Rightarrow (a - \\lambda)(c - \\lambda) - b^2 = 0$<br>\n",
    "\n",
    "$\\Rightarrow ac - \\lambda a - c \\lambda + \\lambda^2 + b^2 = 0$<br>\n",
    "$\\Rightarrow \\lambda^2 - (a+c)\\lambda + ac - b^2$\n",
    "\n",
    "To get eigenvalue $\\lambda$, we can use quadratic equation formula: $x = \\frac{-b + \\sqrt{b^2 - 4ac}}{2a}$\n",
    "\n",
    "$\\Rightarrow \\lambda = \\frac{(a+c) \\pm \\sqrt{(a+c)^2 - 4(ac - b)^2}}{2(1)}$<br>\n",
    "\n",
    "$\\qquad = \\frac{(a+c) \\pm \\sqrt{(a^2 + c^2 + 2ac - 4ac + 4b)^2}}{2}$<br>\n",
    "\n",
    "$\\qquad = \\frac{(a+c) \\pm \\sqrt{(a^2 + c^2 -2ac + 4b)^2}}{2}$<br>\n",
    "\n",
    "$\\qquad = \\frac{(a+c) \\pm \\sqrt{(a - c)^2 + 4b^2}}{2}$<br>\n",
    "\n",
    "$\\Rightarrow \\lambda_{1} = \\frac{(a+c) + \\sqrt{(a - c)^2 + 4b^2}}{2}$, and $\\lambda_{2} = \\frac{(a+c) - \\sqrt{(a - c)^2 + 4b^2}}{2}$<br>\n",
    "\n",
    "Next, getting eigenvector for $\\lambda_{1}$ \n",
    "\n",
    "with $(A - \\lambda I) u = 0$\n",
    "\n",
    "$(A - \\lambda_{1} I) u = \\begin{pmatrix}\n",
    "a - \\lambda_{1} &b\\\\\n",
    "b& c - \\lambda_{1}\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow (a - \\lambda_{1})x + by = 0$<br>\n",
    "and, $bx + (c - \\lambda_{1})y = 0 $\n",
    "\n",
    "$\\Rightarrow x = b$ and $y = \\lambda_{1} -a$\n",
    "\n",
    "Therefore, eigenvector $u = \\begin{pmatrix}\n",
    "b\\\\\n",
    "\\lambda_{1} - a\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "Next, getting eigenvector for $\\lambda_{2}$ \n",
    "\n",
    "with $(A - \\lambda I) u = 0$\n",
    "\n",
    "$(A - \\lambda_{2} I) u = \\begin{pmatrix}\n",
    "a - \\lambda_{2} &b\\\\\n",
    "b& c - \\lambda_{2}\\\\\n",
    "\\end{pmatrix} \\cdot \\begin{pmatrix}\n",
    "x\\\\\n",
    "y\\\\\n",
    "\\end{pmatrix} = \\begin{pmatrix}\n",
    "0\\\\\n",
    "0\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "$\\Rightarrow (a - \\lambda_{2})x + by = 0$<br>\n",
    "and, $bx + (c - \\lambda_{2})y = 0 $\n",
    "\n",
    "$\\Rightarrow x = \\lambda_{2} - c$ and $y = b$\n",
    "\n",
    "Therefore, eigenvector $v = \\begin{pmatrix}\n",
    "\\lambda_{2} - c\\\\\n",
    "b\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Now, normalizing eigenvectors to get orthogonal matrix\n",
    "\n",
    "$\\lVert u \\rVert^2 = \\begin{pmatrix}\n",
    "b\\\\\n",
    "\\lambda_{1}- a\\\\\n",
    "\\end{pmatrix} = b^2 + (\\lambda_{1} - a)^2$\n",
    "\n",
    "$\\Rightarrow \\lVert u \\rVert = \\sqrt{b^2 + (\\lambda_{1} - a)^2}$\n",
    "\n",
    "$\\widehat u = \\frac{1}{\\lVert u \\rVert}u = \\frac{1}{\\sqrt{b^2 + (\\lambda_{1} - a)^2}}\\begin{pmatrix}\n",
    "b\\\\\n",
    "\\lambda_{1}- a\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "\n",
    "$\\lVert v \\rVert^2 = \\begin{pmatrix}\n",
    "\\lambda_{2} - c\\\\\n",
    "b\\\\\n",
    "\\end{pmatrix} = (\\lambda_{2} - c)^2 + b^2$\n",
    "\n",
    "$\\Rightarrow \\lVert v \\rVert = \\sqrt{(\\lambda_{2} - c)^2 + b^2}$\n",
    "\n",
    "$\\Rightarrow v = \\frac{1}{\\lVert v \\rVert}v = \\frac{1}{\\sqrt{(\\lambda_{2} - c)^2 + b^2}} \\begin{pmatrix}\n",
    "\\lambda_{2} -c\\\\\n",
    "b\\\\\n",
    "\\end{pmatrix}$\n",
    "\n",
    "Orthogonal matrix, $Q$ is given by $Q = (\\widehat u \\widehat v)$\n",
    "\n",
    "Therefore, $Q = \\begin{pmatrix}\n",
    "\\frac{b}{\\sqrt{b^2 + (\\lambda_{1} - a)^2}} & \\frac{\\lambda_{2} - c}{(\\lambda_{2} - c)^2 + b^2}\\\\\n",
    "\\frac{\\lambda_{1}-a}{\\sqrt{b^2 + (\\lambda_{1} - a)^2}} & \\frac{b}{(\\lambda_{2} - c)^2 + b^2}\\\\\n",
    "\\end{pmatrix}$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "1109645d",
   "metadata": {},
   "source": [
    "--------------------- "
   ]
  },
  {
   "cell_type": "markdown",
   "id": "ffa42f38",
   "metadata": {},
   "source": [
    "<b> Question 9 </b> \n",
    "\n",
    "Let $A$ be a symmetric invertible matrix. If $Q$ orthogonally diagonalizes the matrix $A$ show that $Q$ also diagonalizes the matrix $A^{-1}$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "d30054c4",
   "metadata": {},
   "source": [
    "<b> Solution </b>\n",
    "\n",
    "We have, $Q$ orthogonally diagonalizing matrix $A$. and we know that a matrix $A$ is orthogonally diagonalizable if is an orthogonal matrix $Q$ such that\n",
    "\n",
    "$Q^TAQ  = D$ where $D$ is a diagonal matrix\n",
    "\n",
    "So, we have<br>\n",
    "$D  = Q^TAQ$\n",
    "\n",
    "Next, taking inverse both side, we have\n",
    "\n",
    "$D^{-1} = (Q^TAQ)^{-1}$<br>\n",
    "$\\qquad = (Q^T)^{-1}A^{-1}Q^{-1}$<br> \n",
    "$\\qquad = (Q^{-1})^{-1}A^{-1}Q^{-1}$<br>\n",
    "$\\qquad = QA^{-1}Q^{-1}$<br> \n",
    "$\\qquad = QA^{-1}Q^T$<br> \n",
    "\n",
    "We can see that matrix $A^{-1}$ is also diagonalizable by orthogonal matrix, $Q$ where $D^{-1}$ is a diagonal matrix."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "de7fac4f",
   "metadata": {},
   "outputs": [],
   "source": []
  }
 ],
 "metadata": {
  "kernelspec": {
   "display_name": "Python 3 (ipykernel)",
   "language": "python",
   "name": "python3"
  },
  "language_info": {
   "codemirror_mode": {
    "name": "ipython",
    "version": 3
   },
   "file_extension": ".py",
   "mimetype": "text/x-python",
   "name": "python",
   "nbconvert_exporter": "python",
   "pygments_lexer": "ipython3",
   "version": "3.9.9"
  }
 },
 "nbformat": 4,
 "nbformat_minor": 5
}