Dimension of centraliser of an $n \times n$ matrix is atleast n.

Question

Let $A$ be an $n \times n$ matrix over a field $\mathbb{F}.$ Then consider the subspace of $M_n(\mathbb{F})$ given by $W=\{ B \in M_n(\mathbb{F}): AB =BA\}.$ Then I want to show that $\text{dim}(W) \geq n.$

What I understand is that if we consider the Rational Canonical form of the matrix then there is an isomorphism between $W$ and the centraliser of the rational canonical form by conjugation. I have seen few answers from stack, but I couldn't understand. If someone give me the key idea it will help me a lot. Thanks.

Answer 1

If you know the rational canonical form, then it is very easy.

The RCF of $A$ is in the form $\tilde{A}=diag(A_1,\cdots,A_k)$ where $A_j\in M_{n_j}(F)$ is cyclic ($n_1+\cdots+n_k=n$), that is, the system $\{I_{n_j},A_j,\cdot,A_j^{n_j-1}\}$ is free. Then $C(A_j)$, the centralizer of $A_j$, has dimension $\geq n_j$.

Note that $U=\{B=diag(B_1,\cdots,B_k);B_jA_j=A_jB_j,j=1,\cdots,k\}$ is a subvector space of $C(\tilde{A})$ and $dim(U)\geq n_1+\cdots+n_k=n$; finally $n\leq dim(C(\tilde{A}))=dim(C(A))$. $\square$