Question

My Determine whether the given matrix is diagonalizable; if so, find a matrix P and a diagonal matrix D such that A - PDP1. (If the matrix is not dlagonalizable, enter DNE in any cell.) T o 1 0 A-1 20 L-1 1 1 [PD] Additional Materials Tutorial Show My Work (optiena) Submit Answer Save Progress Practice Another Version 25

linear algebra

Answer 1

Solution :

$\mathrm{Steps\:}:$

$\begin{pmatrix}0&1&0\\ -1&2&0\\ -1&1&1\end{pmatrix}$

$\mathrm{A\:matrix\:}A\mathrm{\:can\:be\:diagonalized\:if\:there\:exists\:an\:invertible\:matrix\:}P\mathrm{\:and\:diagonal\:matrix\:}D$

$\mathrm{\:such\:that\:}A=PDP^{-1}$

$\mathrm{Eigenvalues\:for\:}\begin{pmatrix}0&1&0\\ -1&2&0\\ -1&1&1\end{pmatrix}:\quad$

$\begin{pmatrix}0&1&0\\ -1&2&0\\ -1&1&1\end{pmatrix}$

$\mathrm{Eigenvalues\:of\:}A\mathrm{\:are\:the\:roots\:of\:the\:characteristic\:equ.\:}\det \left(A-\lambda \:I\right)=0$

$\det \left(\begin{pmatrix}0&1&0\\ -1&2&0\\ -1&1&1\end{pmatrix}-\lambda \begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}\right)$

$=\det \left(\begin{pmatrix}0&1&0\\ -1&2&0\\ -1&1&1\end{pmatrix}-\begin{pmatrix}\lambda &0&0\\ 0&\lambda&0\\ 0&0&\lambda\end{pmatrix}\right)$

$=\det \begin{pmatrix}-\lambda &1&0\\ -1&2-\lambda&0\\ -1&1&1-\lambda\end{pmatrix}$

$\det \begin{pmatrix}-\lambda &1&0\\ -1&2-\lambda&0\\ -1&1&1-\lambda\end{pmatrix}$

$=-\lambda* \det \begin{pmatrix}2-\lambda&0\\ 1&1-\lambda\end{pmatrix}-1*\det \begin{pmatrix}-1&0\\ -1&1-\lambda \end{pmatrix}+0*\det \begin{pmatrix}-1&2-\lambda \\ -1&1\end{pmatrix}$

$=-\lambda* (\left(2-\lambda\right)\left(1-\lambda\right)-0*1)-1*(\left(-1\right)\left(1-\lambda\right)-0* \left(-1\right))+0*(\left(-1\right)*1-\left(2-\lambda\right)\left(-1\right))$

$=-\lambda* (\left(-\lambda+2\right)\left(-\lambda+1\right))-1*(-\left(-\lambda+1\right))-0* \left(-1\right))+0*(-1+2-\lambda)$

$=-\lambda* (\lambda^{2}-1\lambda-2\lambda+2*1)-1*(\lambda-1)+0*(-1+2-\lambda)$

$=-\lambda* (\lambda^2-3\lambda+2)-1*(\lambda-1)$

$=-\lambda ^3+3\lambda^2-2\lambda-\lambda+1$

$=-\lambda ^3+3\lambda^2-3\lambda+1$

$\mathrm{Solve\:}\:-\lambda ^3+3\lambda ^2-3\lambda +1=0:$

$-\lambda ^3+3\lambda ^2-3\lambda +1=0$

$-(\lambda ^3-3\lambda^2+3\lambda-1)=0$

$\mathrm{Apply\:cube\:of\:difference\:rule:\:}a^3-3a^2b+3ab^2-b^3=\left(a-b\right)^3$

$-(\left(\lambda -1\right)\left(\lambda-1\right)\left(\lambda-1\right))=0$

$-\left(\lambda -1\right)^3=0$

$\mathrm{Using\:the\:Zero\:Factor\:Principle:}$

$\mathrm{Solve\:}\:\lambda -1=0:$

$\lambda -1=0$

$\lambda =1$

$\mathrm{The\:final\:solution\:to\:the\:equation\:is:}$

$\lambda =1$

$\mathrm{The\:eigenvalues\:are:}$

$=1$

$\mathrm{Now,\:the\:eigenvectors\:for\:}\begin{pmatrix}0&1&0\\ -1&2&0\\ -1&1&1\end{pmatrix}:$

$\begin{pmatrix}0&1&0\\ -1&2&0\\ -1&1&1\end{pmatrix}$

$\mathrm{Solve\:}\:\left(A-\lambda \:I\right):\:\begin{pmatrix}0&1&0\\ -1&2&0\\ -1&1&1\end{pmatrix}-1\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}$

$\begin{pmatrix}0&1&0\\ -1&2&0\\ -1&1&1\end{pmatrix}-1* \begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}$

$=\begin{pmatrix}0&1&0\\ -1&2&0\\ -1&1&1\end{pmatrix}-\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}$

$=\begin{pmatrix}0-1&1-0&0-0\\ \left(-1\right)-0&2-1&0-0\\ \left(-1\right)-0&1-0&1-1\end{pmatrix}$

$\mathrm{Simplify\:each\:element}$

$=\begin{pmatrix}-1&1&0\\ -1&1&0\\ -1&1&0\end{pmatrix}$

$\mathrm{Reduce\:}\begin{pmatrix}-1&1&0\\ -1&1&0\\ -1&1&0\end{pmatrix}:\quad$

$\begin{pmatrix}-1&1&0\\ -1&1&0\\ -1&1&0\end{pmatrix}$

$\mathrm{Cancel\:leading\:coefficient\:in\:row\:}\:R_2\:\mathrm{\:by\:performing}\:R_2\:\leftarrow \:R_2-1*R_1$

$=\begin{pmatrix}-1&1&0\\ 0&0&0\\ -1&1&0\end{pmatrix}$

$\mathrm{Cancel\:leading\:coefficient\:in\:row\:}\:R_3\:\mathrm{\:by\:performing}\:R_3\:\leftarrow \:R_3-1*R_1$

$=\begin{pmatrix}-1&1&0\\ 0&0&0\\ 0&0&0\end{pmatrix}$

$\mathrm{Multiply\:matrix\:row\:by\:constant:}\:R_1\:\leftarrow \:-1*R_1$

$=\begin{pmatrix}1&-1&0\\ 0&0&0\\ 0&0&0\end{pmatrix}$

$\mathrm{The\:system\:associated\:with\:the\:eigenvalue\:}\lambda =1$

0 0 0/ z

$\mathrm{This\:reduces\:to\:the\:equation}$

$x-y=0$

$\mathrm{Note\:that\:}z\mathrm{\:did\:not\:show\:up\:in\:the\:equations\:which\:means\:}\begin{pmatrix}1&-1&0\\ 0&0&0\\ 0&0&0\end{pmatrix}\begin{pmatrix}0\\ 0\\ z\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\end{pmatrix}\mathrm{\:regardless\:of\:the\:value\:of\:}z$

$\mathrm{Therefore,\:the\:eigenvectors\:associated\:with\:the\:eigenvalue\:}\lambda =1$

$\begin{pmatrix}0\\ 0\\ 1\end{pmatrix}$

$\mathrm{Isolate}$

$x=y$

$\mathrm{Plug\:into\:}\begin{pmatrix}x\\ y\\ z\end{pmatrix}$

$v=\begin{pmatrix}y\\ y\\ 0\end{pmatrix}\space\space\:y\ne \:0$

$\mathrm{Let\:}y=1$

$\begin{pmatrix}1\\ 1\\ 0\end{pmatrix}$

$\mathrm{All\:eigenvectors\:for\:the\:eigenvalue\:}\lambda =1$

$\begin{pmatrix}0\\ 0\\ 1\end{pmatrix},\:\begin{pmatrix}1\\ 1\\ 0\end{pmatrix}$

$\mathrm{To\:allow\:diagonalization,\:the\:number\:of\:Eigenvectors\:must\:equal\:the\:matrix\:dimensions.}$

$\mathrm{There\:are\:}2\mathrm{\:Eigenvectors\:which\:is\:less\:then\:}3\mathrm{\:and\:therefore\:the\:matix\:cannot\:be\:diagonalized}$

$=\mathrm{Not\:diagonizable}$