Assume $A\in M_n(\mathbb{R})$ which means $A$ is an $n\times n$ matrix over field of real numbers. Define $C(A)=\{B\in M_n(\mathbb{R}) \mid AB=BA\}$. It can be shown that $C(A)$ is a vector space over real numbers field. In general, we can show $\text{dim } C(A)\geq n$. In a specific case, assume $A$ is a diagonal matrix that all entries on main diagonal are pairwise different. In this case, what is $\text{dim }C(A)$?
I suspect that dimension of $C(A)$ in this case is equal to $n$. We know all the diagonal matrices commute with this $A$. So we can introduce a basis with elements $E_1,\dots,E_n$ which $E_i$ is a matrix that its $(i,i)$ entry is one and all the other entries are zeros. So the dimension is $n$. But I'm not sure it's a correct answer, because maybe there are other matrices that commute with $A$ that can't be generated with a linear combination of $E_1,\dots,E_n$. Is this basis correct or no?
Please don't use minimal and characteristic polynomials of a matrix. I would be really thankful if you use reasonings as elemtary as I said.
Any help is so much appreciated!
Best Answer
You are correct in your guess that if $A$ is diagonal with distinct diagonal elements, then the linearly independent elements $E_1,\dots,E_n\in C(A)$ span $C(A)$ (hence constitute a basis of $C(A)$).
Indeed, assuming more generally that $$A=\begin{pmatrix} d_1 I_{n_1}& 0 &\dots& 0\\ 0&d_2 I_{n_2}&\dots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\dots&d_pI_{n_p} \end{pmatrix}$$ where $d_1,\dots,d_p$ are pairwise distinct, let us check that $AB=BA$ iff $B$ is of the form $$\begin{pmatrix} C_1& 0 &\dots& 0\\ 0&C_2&\dots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\dots&C_p \end{pmatrix}$$ where each $C_k$ is an $n_k\times n_k$ matrix.
It is clear that every matrix of this form commutes with $A.$ Conversely, if $B\in C(A)$ then $B$ is of this form since for each $k$, $\ker(A-d_kI_n)$ is closed under $B.$