# Matrices commuting with a given $3\times 3$ complex matrix.

Let $$A$$ be a $$3\times 3$$ complex matrix. Let $$C(A)$$ be the vector space of complex matrices that commute with $$A$$. Show that the complex dimension of $$C(A)$$ is at least $$3$$.

I know that this kind of questions has been asked many times on this site. And there is an explicit formula for the dimension of $$C(A)$$ given by Frobenius viewing the matrices $$B$$ that commute with $$A$$ as endomorphisms of $$\mathbb C[\lambda]$$-module.

But I am looking for a more elementary way to show the lower bound of the dimension of $$C(A)$$ is $$3$$.

For example, I have already found that $$\operatorname{Span}\{ I, A \}$$ is a two-dimensional subspace of $$C(A)$$ for $$A\notin\operatorname{Span}\{I\}$$, where $$I$$ is the identity matrix. But how to find another matrix that is linearly independent of $$\operatorname{Span}\{I, A\}$$? Thanks.

• An easy proof with minimal polynomial is given here (take $n=3$), or here. – Dietrich Burde Jul 25 at 13:37
• @DietrichBurde Yes, I have read this post. But I am looking for an explicit matrix that is linear independent of $\operatorname{Span}\{I, A\}$. – Bach Jul 25 at 13:39
• @DietrichBurde But this is supposed to be finished in 20 minutes in an exam, which clearly all the other fancy ways are not good fits. – Bach Jul 25 at 13:47
• @DietrichBurde No, they're not linearly independent. Proof: $A^3 + 2A^2 + A + I = 0$. – Magma Jul 25 at 14:15

By change of basis we can assume that $$A$$ is in Jordan normal form.

Case 1: $$A = \begin{pmatrix}a&0&0\\0&b&0\\0&0&c\end{pmatrix}$$.

Then the matrices $$\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix}, \begin{pmatrix}0&0&0\\0&1&0\\0&0&0\end{pmatrix}, \begin{pmatrix}0&0&0\\0&0&0\\0&0&1\end{pmatrix}$$ are linearly independent and commute with $$A$$.

Case 2: $$A = \begin{pmatrix}a&1&0\\0&a&0\\0&0&b\end{pmatrix}$$.

Then the matrices $$\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix}, \begin{pmatrix}1&0&0\\0&1&0\\0&0&0\end{pmatrix}, \begin{pmatrix}0&0&0\\0&0&0\\0&0&1\end{pmatrix}$$ are linearly independent and commute with $$A$$.

Case 3: $$A = \begin{pmatrix}a&1&0\\0&a&1\\0&0&a\end{pmatrix}$$.

Then the matrices $$\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}, \begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}, \begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}$$ are linearly independent and commute with $$A$$.

More generally, for each Jordan block of size $$k\times k$$ we can quickly find $$k$$ independent commuting matrices $$A_i$$ by setting the entries on the $$(i-1)$$-th upper off-diagonal to $$1$$ and all other entries to $$0$$. Then for any matrix $$A$$ in Jordan form, you can construct commuting matrices whose nonzero entries are an off-diagonal of ones spanning the location of a single Jordan block of $$A$$.

• I have to say that this pattern is very appealing! I have never been aware of this fact before. Thank you! – Bach Jul 25 at 14:26