Diagonalization of $ A=\left(\begin{smallmatrix} a & a & a \\ a & a & a \\ a & a & a \end{smallmatrix}\right) $

Question

Consider the matrix: $$ A=\left(\begin{matrix} a & a & a \\ a & a & a \\ a & a & a \end{matrix}\right) $$ How to prove whether $A$ is diagonalisable or not. And how to diagonalize it.

Answer 1

To diagonalize, first notice that $$A\begin{bmatrix}1\\1\\1\end{bmatrix}=\begin{bmatrix}3a\\3a\\3a\end{bmatrix}$$ so $$u:=\begin{bmatrix}1\\1\\1\end{bmatrix}$$ is an eigenvector of the $3a$ eigenvalue. To find eigenvectors $v,w$ of the $0$ eigenvalue, simply reduce $A$ to its RREF and determine a basis for the kernel. Then we can diagonalize $A$ as $A=P^{-1}DP$ with $$D:=\begin{bmatrix}3a&0&0\\0&0&0\\0&0&0\end{bmatrix}$$ and $P:=\begin{bmatrix}u&v&w\end{bmatrix}$.