I need to diagonalize this matrix but I'm not sure it can be

Question

This is the matrix I need to diagonalize: $A=\left[\begin{matrix}3&2\\0&3\end{matrix}\right]$. So I found the eigenvalue by taking the determinant of $(A-\lambda I)$ and solving for $\lambda=0$ and found that my eigenvalue is $3$. Then I subtracted $A-3I=\left[\begin{matrix}0&2\\0&0\end{matrix}\right]$. Now my question is whether this is diagonalizable? I want to say it's not because there's only $1$ pivot column and thus the basis wouldn't fill $\mathbb R^2$. But for some reason I feel that my conclusion is correct and this is diagonalizable. Am I doing this right?

Answer 1

The geometric multiplicity of the eigenvalue(s) is $1$, which is less than the algebraic multiplicity, which is $2$. By this, we mean there is only one eigenvector corresponding to the eigenvalue $3$, the eigenvector being $\begin{bmatrix}1 \\ 0\end{bmatrix}$.

Hence, the matrix is not diagonalizable. This is a standard example in matrix algebra for a non-diagonalizable matrix.