Question

Consider the linear transformation T: Rn → Rn whose matrix A relative to the standard basis is given. A 2 2 (a) Find the eigenvalues of A. (Enter your answers from smallest to largest.) (A1, A2) -1 5 (b) Find a basis for each of the corresponding eigenspaces (c) Find the matrix A' for Trelative to the basis B', where B' is made up of the basis vectors found in part (b)

Answer 1

Solution : ( a )

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

$\det \left(\begin{pmatrix}2&2\\ -1&5\end{pmatrix}-\lambda \begin{pmatrix}1&0\\ 0&1\end{pmatrix}\right)$

$= \det \left ( \begin{pmatrix}2&2\\ -1&5\end{pmatrix}-\begin{pmatrix}\lambda &0\\ 0&\lambda \end{pmatrix} \right )$

$= \det \begin{pmatrix}2-\lambda &2-0\\ \left(-1\right)-0&5-\lambda \end{pmatrix}$

$= \det \begin{pmatrix}2-\lambda &2\\ -1&5-\lambda \end{pmatrix}$

$\det \begin{pmatrix}2-\lambda &2\\ -1&5-\lambda \end{pmatrix}$

$=\left(2-\lambda \right)\left(5-\lambda \right)-2\left(-1\right)$

$=\left(2-\lambda \right)\left(5-\lambda \right)+2*1$

$= 2*5+2\left(-\lambda \right)+\left(-\lambda \right)*5+\left(-\lambda \right)\left(-\lambda \right) + 2$

$= 2*5 - 2\lambda -5\lambda +\lambda \lambda + 2$

$= 10 - 7\lambda + \lambda ^2 + 2$

$= \lambda ^2-7\lambda +12$

$\mathrm{Solve\: \:}\lambda^2-7\lambda +12 = 0 :$

$\lambda^2-7\lambda +12 = 0$

$\left(\lambda^2-3\lambda \right)+\left(-4\lambda +12\right) = 0$

$\lambda \left(\lambda-3\right)-4\left(\lambda-3\right) = 0$

$\left(\lambda -3\right)\left(\lambda -4\right)=0$

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

$\mathrm{Solve\:}\:\lambda-3=0:\quad \lambda=3$

$\mathrm{Solve\:}\:\lambda-4=0:\quad \lambda=4$

$\mathrm{The\:final\:solutions\:to\:the\:quadratic\:equation\:are:}$

$\lambda_{1}=3,\:\lambda_{2}=4$

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

$(\lambda _{1},\:\:\lambda _{2})=3,\:4$

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Solution : ( b )

$\mathrm{Eigenvectors\:for\:}\lambda_{1} =3:\quad$

$\mathrm{Solve\:}\:\left(A-\lambda_{1} \:I\right):\:\begin{pmatrix}2&2\\ -1&5\end{pmatrix}-3\begin{pmatrix}1&0\\ 0&1\end{pmatrix}$

$\begin{pmatrix}2&2\\ -1&5\end{pmatrix}-3\begin{pmatrix}1&0\\ 0&1\end{pmatrix}$

$=\begin{pmatrix}2&2\\ -1&5\end{pmatrix}-\begin{pmatrix}3&0\\ 0&3\end{pmatrix}$

$=\begin{pmatrix}2-3&2-0\\ \left(-1\right)-0&5-3\end{pmatrix}$

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

$\mathrm{Reduce\:}\begin{pmatrix}-1&2\\ -1&2\end{pmatrix}:\quad$

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

$\mathrm{Reduce\:matrix\:to\:row\:echelon\:form}\:$

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

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

$\mathrm{Reduce\:matrix\:to\:reduced\:row\:echelon\:form}\:$

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

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

$\mathrm{The\:system\:associated\:with\:the\:eigenvalue\:}\lambda =3$

$\left(A-3I\right)\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}1&-2\\ 0&0\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix}$

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

$x-2y=0$

$\mathrm{Isolate}$

$x=2y$

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

$v=\begin{pmatrix}2y\\ y\end{pmatrix}\space\space\:y\ne \:0$

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

$\begin{pmatrix}2\\ 1\end{pmatrix}$

$\mathrm{and}$

$\mathrm{Eigenvectors\:for\:}\lambda_{2}=4:\quad$

$\mathrm{Solve\:}\:\left(A-\lambda_{2} \:I\right):\:\begin{pmatrix}2&2\\ -1&5\end{pmatrix}-4\begin{pmatrix}1&0\\ 0&1\end{pmatrix}$

$\begin{pmatrix}2&2\\ -1&5\end{pmatrix}-4\begin{pmatrix}1&0\\ 0&1\end{pmatrix}$

$=\begin{pmatrix}2&2\\ -1&5\end{pmatrix}-\begin{pmatrix}4&0\\ 0&4\end{pmatrix}$

$=\begin{pmatrix}2-4&2-0\\ \left(-1\right)-0&5-4\end{pmatrix}$

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

$\mathrm{Reduce\:}\begin{pmatrix}-2&2\\ -1&1\end{pmatrix}:\quad$

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

$\mathrm{Reduce\:matrix\:to\:row\:echelon\:form}\:$

$\mathrm{Cancel\:leading\:coefficient\:in\:row\:}\:R_2\:\mathrm{\:by\:performing}\:R_2\:\leftarrow \:R_2-\frac{1}{2}*R_1$

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

$\mathrm{Reduce\:matrix\:to\:reduced\:row\:echelon\:form}\:$

$\mathrm{Multiply\:matrix\:row\:by\:constant:}\:R_1\:\leftarrow \:-\frac{1}{2}* R_1$

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

$\mathrm{The\:system\:associated\:with\:the\:eigenvalue\:}\lambda =4$

$\left(A-4I\right)\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}1&-1\\ 0&0\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix}$

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

$x-y=0$

$\mathrm{Isolate}$

$x=y$

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

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

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

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

$\mathrm{Hence,}$

$\mathrm{The\:vector\:B_{1}=}\mathrm{\begin{pmatrix}2\\ 1\end{pmatrix}\:\: is \:a\: basis\: for\: the\: eigenspace\: corresponding\: to\:\lambda_{1} =3\:}.$

$\mathrm{and}$

$\mathrm{The\:vector\:B_{2}=}\mathrm{\begin{pmatrix}1\\ 1\end{pmatrix}\:\: is \:a\: basis\: for\: the\: eigenspace\: corresponding\: to\:\lambda_{2} =4\:}.$