Let \begin{equation} A = \begin{bmatrix}8 & -9 & -2 \\ 6 & -7 & -2 \\ -6 & 9 & 4 \end{bmatrix} \end{equation}
Let \begin{equation} V(A) = \{X \in M_{3}(R) : AX = XA \} \end{equation}
where $M_{3}(R)$ is the vector space of $3 \times 3$ real matrices.
Calculate the dimension of $V(A)$.
So far, I have found that A has eigenvalue $\lambda = 1$ with corresponding eigenvector $\vec{v} = \begin{bmatrix}-1 \\ -1 \\ 1 \end{bmatrix}$, and $\lambda = 2$ with corresponding eigenvectors $\vec{v_{1}} = \begin{bmatrix}\frac{3}{2} \\ 1 \\ 0 \end{bmatrix}$ and $\vec{v_{2}} = \begin{bmatrix}\frac{1}{2} \\ 0 \\ 1 \end{bmatrix}$.
I do not know what to do about this centraliser though. All I can tell is that obviously $I,A,A^{-1}$ will all commute with A, so that would bring the dimension up to at least 2? And I understand that the maximumm is 9, but that's about it.
Must you always compute the basis for such a set to find its dimension? Or is there a simple way to predict what the dimension of $V(A)$ will be without doing so? Thanks.
Hint Usually to find the centraliser of a matrix $A$ you have to diagonalise the matrix (if it's possible) and write $A = PDP^{-1}$. Since :
$$ X \mapsto P^{-1}XP $$
is a bijection it then suffices to describe the set :
$$ \{ Y \in \mathcal M_n(\mathbb R) | DY = YD \} $$
with $D$ the diagonal matrix with the eigenvalues of $A$.