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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.

Bach
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    An easy proof with minimal polynomial is given here (take $n=3$), or here. – Dietrich Burde Jul 25 '19 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 '19 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 '19 at 13:47
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    @DietrichBurde No, they're not linearly independent. Proof: $A^3 + 2A^2 + A + I = 0$. – Magma Jul 25 '19 at 14:15

1 Answers1

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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 $A$.

Magma
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  • I have to say that this pattern is very appealing! I have never been aware of this fact before. Thank you! – Bach Jul 25 '19 at 14:26