constructing bases in $\mathbb R^2$ to find diagonalization of a matrix

Question

Let A = $$ \begin{pmatrix} 1 & 1 \\ 1 & 1 \\ \end{pmatrix} $$

Construct bases $\beta,\gamma $ of $\mathbb{R}^2$ such that $[A]_{\beta}^{\gamma}$ is a diagonal matrix

question

Answer 1

The bases $\{\beta,\gamma\}$ will be two eigen vectors corresponding to the two eigen values 0 and 2 of $A$, which are respectively $\beta=(1,-1)^t$ and $\gamma=(1,1)^t$. Note that $A\beta=0$ and $A\gamma=2\gamma$.

Hence, with respect to $\{\beta,\gamma\}$,$$A= \begin{pmatrix} 0 & 0 \\ 0 & 2 \\ \end{pmatrix} $$

Answer 2

The characteristic polynomial is $(x-1)^2-1=0\implies x=0,2$ are the e-values.

We need a basis of eigenvectors. $Ax=0\implies x=s\begin{pmatrix}1\\-1\end{pmatrix}$. And, $Ax=2x\implies x=t\begin{pmatrix}1\\1\end{pmatrix}$.

$\therefore \gamma =\beta=\{\begin{pmatrix}1\\-1\end{pmatrix},\begin{pmatrix}1\\1\end{pmatrix}\}$ will do.

In fact, $[A]_{\beta}^{\gamma}=\begin{pmatrix}0&0\\0&2\end{pmatrix}=\begin{pmatrix}1&1\\-1&1\end{pmatrix}^{-1}A \begin{pmatrix}1&1\\-1&1\end{pmatrix}$.