# Dimension of space of $4\times 4$ matrices that centralize a given matrix

Problem description:

Let A be an $4\times 4$ complex matrix, determine the largest possible dimension of the subspace $S_A=\{B \in M_n (C)|AB=BA\}$.

My answer is 10. Because all commuting matrices are triangulable, so the largest number of the unit matrices $E_{ij}$ equal to the basis that spans an upper triangular matrix.

Is my answer right? Or can you help to offer a proof.

Thanks.

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It's not 10 in the special case that $A$ is the identity matrix, so maybe you need to provide more details of your line of thought? – rschwieb May 2 '12 at 19:29
Over $\mathbb{C}$, all matrices are triangularizable; your argument doesn't make much sense. What do $E_{ij}$ have to do with anything? Note that $E_{ij}A$ is the matrix whose $i$th row is the $j$th row of $A$, and $0$s elsewhere, whereas $AE_{ij}$ is the matrix whose $j$th column is the $i$th column of $A$ and $0$s elsewhere. There are very few matrices $A$ for which $AE_{ij}=E_{ij}A$! – Arturo Magidin May 2 '12 at 19:30

This is a more generic answer...

Suppose $A$ has distinct eigenvalues $\lambda_1,...,\lambda_n$, with corresponding right and left eigenvectors $v_1,...,v_n$ and $u_1^*,...,u_n^*$ respectively. Then it is easy to see that the linear operator $L_A(B) = AB-BA$ has $n^2$ eigenvalues $\mu_{ij} = \lambda_i-\lambda_j$ with corresponding linearly independent eigenvectors $v_i u_j^*$. The kernel of $L_A$ corresponds to the span of the eigenvectors corresponding to the eigenvalues $\mu_{ii} = 0$. Consequently, $\dim \ker L_A = n$.

Since $S_A = \ker L_A$, we have that, in a generic sense, $\dim S_A = n$.

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great answer, though it is beyond my level. thanks. – David May 3 '12 at 4:36

If I understand you correctly you are looking for the bigest possible dimension. This is attained for $A=E_4$ with $S_A=M_4(\mathbb C)$ which has dimension $16$. Here $E_4$ denotes the identity matrix. Alternatively we could also choose $A=0$ and still get $S_A=M_4(\mathbb C)$.

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Surely you mean for $A=0$? (I'm not exactly what you mean by $E_4$.) – copper.hat May 2 '12 at 19:43
how do you reach the answer of E4? what do you mean by that? – David May 2 '12 at 19:47
@user30114 He probably means, as I mentioned above, the identity matrix. Everything commutes with it. – rschwieb May 2 '12 at 19:54