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For a square matrix $A\in F^{n\times n}$ over a field $F$, define the commutator subspace $C_A = \{ B\in F^{n\times n} \vert AB = BA\}$ of matrices which commute with $A$. This other question by RiaD asks for a proof that $\dim C_A\geq n$ for any $A$. The answer by Johannes Kloos uses Jordan Normal Form, and so yields the same result over any algebraically closed field $F$ in place of $\mathbb{C}$.

Now let's write $C_{F,A}$ in place of $C_A$ to keep track of the field. Note that $C_{F,A}$ is the kernel of the linear map $B\mapsto AB-BA$. The dimension of a kernel doesn't change in a field extension because row reduction proceeds in the same way regardless of field, so if $K$ is a field extension of $F$ then $\dim_F C_{F,A} = \dim_K C_{K,A}$. Since any field $F$ is contained in an algebraically closed field $K$ we can combine these facts to get $\dim_F C_{F,A} = \dim_K C_{K,A}\geq n$, so the original result holds for all fields $F$, not just algebraically closed fields.

The question: is there a purely linear algebraic proof of this? By this I mean one which works for all fields and does not involve passing to a field extension.

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Very nice question. –  joriki Mar 21 '12 at 21:48

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Sure. The hard step is to prove that any matrix $A$ can be written (up to conjugation) as the direct sum of matrices $A_i$ whose characteristic and minimal polynomials coincide. Once you know this, it suffices to prove the statement for each $A_i$, but it's obvious because in this case the dimension of $\text{span}(1, A_i, A_i^2, ... A_i^{k-1})$ (where $A_i$ is a $k \times k$ matrix) is $k$ and this is clearly contained in $C_{A_i}$.

The hard step follows from e.g. the structure theorem for finitely-generated modules over a PID (here $F[x]$) which I think is linear algebra just as much as Jordan decomposition is. Anyway, it tells you that $F^n$, as a $F[x]$-module with $x$ acting by $A$, decomposes as a direct sum $$\bigoplus_i F[x]/p_i(x)^{d_i}$$

where $p_i$ is an irreducible polynomial. The restriction of $A$ to each summand has the property that the characteristic and minimal polynomials coincide and we are done.

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How do you mean "which I think is linear algebra just as much as Jordan decomposition is"? I thought the point of the question was that Jordan decomposition only works over an algebraically closed field and the desired proof shouldn't take a detour via algebraically closed fields. –  joriki Mar 21 '12 at 22:08
    
@joriki: I mean two things. First, that Jordan decomposition is also a specialization of the structure theorem, just to the special case that $F$ is algebraically closed so all the irreducible polynomials in $F[x]$ are linear. Second, that the structure theorem and its proof (say via Smith normal form) amount to "linear algebra over PIDs" which is really not much harder than linear algebra over fields. –  Qiaochu Yuan Mar 21 '12 at 22:10
    
Good point. The part I was forgetting was that companion matrices have matching characteristic and minimal polynomials. So rational canonical form works just as well as Jordan normal form and avoids going to a larger field. Although you answered the question, I'll hold off giving you the check mark for now to see if anyone comes up with a proof that avoids the structure theorem for PIDs. –  Noah Stein Mar 22 '12 at 0:02
    
@Noah: I am not sure you can avoid the structure theorem. It is possible (although I am not sure how I would show this) that the "hard step" above is equivalent to the structure theorem, and I don't know how I'd go about proving this without the hard step. –  Qiaochu Yuan Mar 22 '12 at 1:06

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