How to use the basis in order to diagonalize the matrix

Question

Find a basis of eigenvectors of $\mathbb{R}^2$ for the matrix: $$A=\begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}$$ Use that basis in order to diagonalize the above matrix.

My try:

I found the basis of $\begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}=2\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}-1\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$

So, the required basis are $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ and $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$

I know that to diagonalize the matrix, I need $P^{-1}AP$

My question is how should I use these basis for the diagonalization.

Answer 1

My guess is that what you're after is a basis of eigenvalues. One such basis would be $\bigl((1,-1),(1,1)\bigr)$. And, since the correspondent eigenvalues are $3$ and $1$, you have$$\begin{pmatrix}1&1\\-1&1\end{pmatrix}^{-1}\begin{pmatrix}2&-1\\1&2\end{pmatrix}\begin{pmatrix}1&1\\-1&1\end{pmatrix}=\begin{pmatrix}3&0\\0&1\end{pmatrix}.$$

Answer 2

The columns of $P$ will be the eigenvectors of $A$, and then this will give $P^{-1}AP$ as a diagonal matrix whose $i$-th diagonal entry is the eigenvalue of $A$ corresponding to the $i$-th column of $P$ (which is an eigenvector of $A$).

Note: if the eigenvectors of $A$ are linearly dependent, then $P^{-1}$ does not exist, and in fact $A$ is not diagonalizable in this case.