# Dimension of subspace of commuting matrices

I was hoping for some confirmation of a proof to the following preliminary exam question:

Fix an $n \times n$ matrix $A$ with entries in an algebraically closed field $k$. Let $C$ be the space of $n \times n$ matrices over $k$ that commute with $A$. Observe that $C$ is a vector space over $k$. Show that $\dim C \ge n$ and the equality holds if and only if the characteristic polynomial of $A$ equals the minimal polynomial of $A$.

Now there are a lot of questions and answers related to this on MSE but my approach is fairly different from anything I saw, hence the request for verification.

Observe that $n \times n$ matrices have a natural action on $V=k^n$, and we obtain a $k[x]$-module structure on $V$ by letting $x$ act by $A$.

Note that any matrix $B$ commuting with $A$ defines a linear transformation of $V$ which commutes with the action of $x$. In other words, $B \in \text{End}_{k[x]} V$.

Suppose momentarily that $V = k[x]/(a(x))$, where $a$ has degree $n$. Then $V$ is cyclic as a $k[x]$-module, so as a $k[x]$-homomorphism, $B$ can be described by where it maps the generator $1 \in k[x]/(a(x))$. Moreover this is described by the image of a polynomial in the quotient:

$$B \cdot 1 = b_0 + b_1x + b_2 x^2 + \dots + b_{n-1} x^{n-1} \in k[x]/(a(x)).$$

For the sake of clarity, when we describe $B$ as a linear transformation (that is, to give it as a map of vector spaces, rather than as a map of $k[x]$-modules), we define it on the basis elements $1,x,x^2,\dots,x^{n-1}$ by

$$B \cdot x^i = (B \cdot 1) x^i \mod a(x) ,$$

so if

$$a(x) = a_0+a_1x+\dots +a_{n-1}x^{n-1}+x^n$$

then for example

$$B \cdot x = b_0x+b_1x^2+ \dots + b_{n-1}x^n = -a_0b_{n-1}+(b_0-a_1b_{n-1})x+(b_1-a_2b_{n-1})x^2+\dots + (b_{n-2}-a_{n-1}b_{n-1})x^{n-1} .$$

Now, referring to the description of $B$ above, we observe that any choice of $(b_0,b_1,b_2,\dots,b_{n-1}) \in k^n$ yields a well-defined $k[x]$-module homomorphism; in other words, we see that in this special case, $\dim C = \dim \text{End}_{k[x]}=n$.

In general, because $k[x]$ is a PID, we obtain an isomorphism

$$V \cong k[x]/(a_1(x)) \oplus \cdots \oplus k[x]/(a_m(x))$$

by the structure theorem (with no free part because $V$ is finite dimensional), such that each $a_i$ is monic with degree at least $1$, and $a_1 | a_2 | \dots |a_m$.

Now on each summand, define $B_i$ by any choice of $\deg a_i(x)$ elements of $k^n$, as described in the special case above. Then $B=B_1\oplus B_2 \oplus \dots \oplus B_m$ commutes with $A$ (after conjugating back to the original basis), and because the product of invariant factors is the characteristic polynomial, the sum of these degrees is $n$, so we have produced an $n$-dimensional subspace $C'$ of $C$, proving the first part.

On the other hand, if the minimal polynomial is equal to the characteristic polynomial, the only invariant factor is the minimal polynomial, so $V$ is cyclic and we are in the special case above, where we proved $\dim C = n$.

Finally, suppose the minimal factor is not equal to the characteristic polynomial. Then there are at least two invariant factors $a_1|a_2$. But then there is a well-defined $k[x]$-module homomorphism

$$B: k[x]/(a_2(x)) \to k[x]/(a_1(x))$$

defined by extending $1 \mapsto 1$. Then extending $B$ by zero on the other summands of $V$, we obtain a $k[x]$-module homomorphism which cannot be decomposed into a direct sum of homomorphisms on the summands. Thus $B$ is not contained in the subspace $C'$ defined above, so $\dim C > \dim C'=n$.

This completes the proof.

Edit: Note, alarmingly, that I didn't use the fact that $k$ is algebraically closed! This in particular has me worried.

• You should be aware that the minimal polynomial does not change when the field is extended. Requires its own proof – Will Jagy Dec 4 '15 at 3:55
• – Will Jagy Dec 4 '15 at 4:06
• @WillJagy I did find most of those, but they do not quite verify this particular proof. How does the fact that the minimal polynomial doesn't change when extending the field (directly) relate to the problem? – Max Dec 4 '15 at 4:28
• Mostly that you commented on not having used the assumption that $k$ was algebraically closed. You can take $k$ the smallest field containing the matrix entries and get the same result – Will Jagy Dec 4 '15 at 4:31
• Well, I know that the minimal polynomial doesn't change, but over an algebraically closed field, linear transformations have a particularly nice description (Jordan canonical form), which I expected to use. – Max Dec 4 '15 at 4:34