Question

Determine the value of a so that \( \lambda = 2 \) is an eigenvalue of 

\( A=\begin{pmatrix}1&-1&0\\ a&1&1\\ 0&1+a&3\end{pmatrix} \)

Then show that A is diagonallizable and diagonalize it. 

Answer 1

Solution

we have \( A-2I=\begin{pmatrix}-1&-1&0\\ a&-1&1\\ 0&1+a&3\end{pmatrix} \)

Then if \( \lambda =2\in SP(A) \)

\( \implies |A-2I|=\begin{vmatrix}-1&-1&0\\ \:a&-1&1\\ \:0&1+a&3\end{vmatrix}=0 \) Thus, \( a=-1 \)

we have \( A=\begin{pmatrix}1&-1&0\\ -1&1&1\\ 0&0&3\end{pmatrix} \)

\( \implies P(\lambda)=|A-\lambda I|=\begin{vmatrix}1-\lambda &-1&0\\ \:-1&1-\lambda \:&1\\ \:0&0&3-\lambda \:\end{vmatrix} \)

                    \( =\bigg(3-\lambda\bigg)\bigg[\bigg(1-\lambda\bigg)^2-1\bigg] \)

\( \implies P(\lambda)=\bigg(3-\lambda\bigg)\bigg(1-2\lambda+\lambda^2-1\bigg) \)

                     \( =\lambda\bigg(\lambda-2\bigg)\bigg(3-\lambda\bigg) \)

since, \( A\in M_3(\mathbb{R}) \) and \( P(\lambda)=\lambda\bigg(\lambda-2\bigg)\bigg(3-\lambda\bigg) \)

\( \color{red}Note : Theorem21 \)

(1) The characteristic polynomial \( P_L(\lambda) \) splits over \( \mathbb{K} \)

(2) Every eigenvalue \( \lambda \) of L \( am(\lambda)=gm(\lambda) \)

By Theorem 21\( \implies am(\lambda)=gm(\lambda)=1 ,\forall \lambda \iff SP(\lambda) \)

\( \bullet \) Find  \( E_0 \)

\( \implies A-(0)I= \begin{pmatrix}1&-1&0\\ -1&1&1\\ 0&0&3\end{pmatrix}\sim \begin{pmatrix}0&0&1\\ 1&-1&0\\ 0&0&0\end{pmatrix}\hspace{2mm} \)

Then \( E_0=span\left\{\begin{pmatrix}1\\ 1\\ 0\end{pmatrix}\right\} \)

\( \bullet \) Find \( E_2 \)

\( \implies A-2I=\begin{pmatrix}-1&-1&0\\ -1&-1&1\\ 0&0&1\end{pmatrix}\sim \begin{pmatrix}-1&-1&0\\ 0&0&1\\ 0&0&0\end{pmatrix} \)

Then \( E_2=span\left\{\begin{pmatrix}1\\ -1\\ 0\end{pmatrix}\right\} \)

\( \bullet \) Find \( E_3 \) 

\( \implies A-3I=\begin{pmatrix}-2&-1&0\\ -1&-2&1\\ 0&0&0\end{pmatrix}\sim \begin{pmatrix}2&1&0\\ 0&3&1\\ 0&0&0\end{pmatrix} \)

Then \( E_3=span\left\{\begin{pmatrix}-1\\ 2\\ 3\end{pmatrix}\right\} \)

since. \( dimE_0+dimE_2+dimE_3=3=\mathbb{R} \)

A is diagonalizable in \( \mathbb{R} \)

Therefore. \( A=PDP^{-1} \) where \( D=\begin{pmatrix}0&0&0\\ 0&2&0\\ 0&0&3\end{pmatrix},P=\begin{pmatrix}1&1&-1\\ 1&-1&2\\ 0&0&3\end{pmatrix} \)

Answer

Therefore.

\( A=PDP^{-1} \) where  \( D=\begin{pmatrix}0&0&0\\ 0&2&0\\ 0&0&3\end{pmatrix} \) and \( P=\begin{pmatrix}1&1&-1\\ 1&-1&2\\ 0&0&3\end{pmatrix} \)