91Ó°ÊÓ

We solve:

$$\begin{gathered} det\left(A-\mathrm{λ}l\right)=0 \\ \left|\begin{array}{cc}1-\mathrm{λ}& 1\\ 1& 1-\mathrm{λ}\end{array}\right|=0 \\ {\left(1-\mathrm{λ}\right)}^2-1=0 \\ {\mathrm{λ}}^2-2\mathrm{λ}=0 \\ \mathrm{λ}\left(\mathrm{λ}-2\right)=0 \\ {\mathrm{λ}}_1=0,{\mathrm{λ}}_2=2 \end{gathered}$$

For$\mathrm{λ}=0$, we solve:

$$\begin{gathered} \left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right] \\ {x}_1+y_1=0 \end{gathered}$$

We can choose an eigenvector:

$v_1=\left[\begin{array}{c}1\\ -1\end{array}\right]$

For$\mathrm{λ}=2$, we solve:

$$\begin{gathered} \left[\begin{array}{cc}-1& 1\\ 1& -1\end{array}\right]\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right] \\ {x}_1-y_1=0 \end{gathered}$$

We can choose an eigenvector:

$v_2=\left[\begin{array}{c}1\\ -1\end{array}\right]$

Now, localid="1659533233518" $\left\{v_1,v_2\right\}$is an eigenbasis for $R^2$, therefore the diagonalization of A in the eigenbasis is $\left[\begin{array}{cc}0& 0\\ 0& 2\end{array}\right]$_.

Hence the final answer is $\left[\begin{array}{cc}0& 0\\ 0& 2\end{array}\right]$.