Question

Find all eigenvalues and their corresponding eigenspaces of the following matrix. B=\( \begin{pmatrix}2&-3&1\\ 1&-2&1\\ 1&-3&2\end{pmatrix} \)

Answer 1

Solution

B=\( \begin{pmatrix}2&-3&1\\ 1&-2&1\\ 1&-3&2\end{pmatrix} \)  The charateristic polynomial of B is.

\( P_B(\lambda)=|B-\lambda I|=\begin{vmatrix}2-\lambda &-3&1\\ 1&-2-\lambda &1\\ 1&-3&2-\lambda \end{vmatrix}=-\lambda^3+2\lambda^2-\lambda \)

\( set \hspace{2mm}P_B(\lambda)=0\iff -\lambda^3+2\lambda^2-\lambda=0 \implies \lambda_1=0,\lambda_2=1 \hspace{2mm}is\hspace{2mm} eigenvalue. \)\( \implies eigenspaces \hspace{2mm}E_\lambda = \left\{{x\in \mathbb{R}^3|(B-\lambda I)x=0}\right\} \)

\( For\hspace{2mm}\lambda_1=0\implies B-(0)I=B=\begin{pmatrix}2&-3&1\\ 1&-2&1\\ 1&-3&2\end{pmatrix} \hspace{3mm} \sim\hspace{3mm} \begin{pmatrix}1&-3&2\\ 1&-2&1\\ 2&-3&1\end{pmatrix}\hspace{3mm} \sim \hspace{3mm} \begin{pmatrix}1&0&-1\\ 0&1&-1\\ 0&0&0\end{pmatrix} \)\( Let \hspace{2mm}t=x_3,t\in \mathbb{R}\implies x_2=t,x_1=t \)

\( Then,\hspace{2mm}x=\bigg(t,t,t\bigg)=t\bigg(1,1,1\bigg)\implies E_0=\begin{pmatrix}t\\ t\\ t\end{pmatrix} \hspace{2mm} is\hspace{2mm} eigenspace. \)

\( Therefore. E_0=span\bigg(1,1,1\bigg) \)\( For\hspace{2mm} \lambda_2=1\implies B-I=\begin{pmatrix}1&-3&1\\ 1&-3&1\\ 1&-3&1\end{pmatrix} \hspace{2mm}\sim \hspace{2mm}\begin{pmatrix}1&-3&1\\ 0&0&0\\ 0&0&0\end{pmatrix}\hspace{2mm}\hspace{2mm}Let. \)

d\( Let.\hspace{2mm} x_2=t,x_3=s\implies x_1=3x_2-x_3=3t-s \)

\( Then, \hspace{2mm} x=\bigg(3t-s,t,s\bigg)\implies E_0=\begin{pmatrix}3t-s\\ t\\ s\end{pmatrix}=span{\left\{\begin{pmatrix}3\\ 1\\ 0\end{pmatrix},\begin{pmatrix}-1\\ 0\\ 1\end{pmatrix}\right\}} \)

 

 

Answer : 

Therefore

Therefore. \( E_0=span\bigg(1,1,1\bigg) \)

\( Then, \hspace{2mm} x=\bigg(3t-s,t,s\bigg)\implies E_0=\begin{pmatrix}3t-s\\ t\\ s\end{pmatrix}=span{\left\{\begin{pmatrix}3\\ 1\\ 0\end{pmatrix},\begin{pmatrix}-1\\ 0\\ 1\end{pmatrix}\right\}} \)