Consider the matrix $A = \begin{bmatrix}3 &0\\ 0& 3\end{bmatrix}$. To determine whether it can be diagonalised, I have found eigvalues $\lambda_1,\lambda_2$ which are both $= 3$, but then I get an eigvec of $\begin{bmatrix}0\\0\end{bmatrix}$. The solution says it's diagonalisable but I don't see how that is true because the eigvec can't be a 0 eigvec.
edit the original question the $A$ matrix was accidentally $A = \begin{bmatrix}3 &0\\ 3& 0\end{bmatrix}$ which gave the answer below which could also be useful if you are new to learning this.
$x^2-3x=0$ is the characteristic polynomial of $\begin{bmatrix}3&0\\3&0 \end{bmatrix}$ yielding the roots $x=0,3$.
To find the eigenvectors, we find the null space of the two matrices formed by subtracting eigenvalues from the diagonal entries: $\begin{bmatrix}3&0\\3&0 \end{bmatrix}$ with null space spanned by $\begin{bmatrix}0 \\ 1\end{bmatrix}$.
and the second matrix: $\begin{bmatrix}0&0\\3&-3 \end{bmatrix}$, with nullspace spanned by $\begin{bmatrix}1\\ -1 \end{bmatrix}$.
Yielding two linearily independent eigenvectors and telling you that your matrix is indeed diagonalizable.