Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The following is an interesting problem from Linear Algebra 2nd Ed - Hoffman & Kunze (3.5 Q17).

Let $W$ be the subspace spanned by the commutators of $M_{n\times n}\left(F\right)$: $$C=\left[A, B\right] = AB-BA$$ Prove that $W$ is exactly the subspace of matrices with zero trace.

Assuming this is true, one can construct $n^2 - 1$ linearly independent matrices, in particular $$[e_{i,n}, e_{n,i}]\ \text{for $1\le i\le n-1$}$$ $$[e_{i,n}, e_{j,n}]\ \text{for $i\neq j$}$$ where $e_{i,j}$ are the standard basis with $0$ entry everywhere except row $i$ column $j$ which span the space of traceless matrices.

However, I have trouble showing (or rather, believing, since this fact seems to be given) that the set of commutators form a subspace. In particular, I am having difficulty showing that the set is closed under addition. Can anyone shed some light?

share|improve this question

2 Answers 2

up vote 9 down vote accepted

The set of matrix commutators is in fact a subspace, as every commutator has trace zero (fairly easy to prove) and every matrix with trace zero is a commutator (stated here but I know of no elementary proof), and the set of traceless matrices is clearly closed under linear combinations.

However, the problem is talking about the subspace spanned by the set of matrix commutators, which means the set of linear combinations of matrix commutators. This is by definition a subspace. This is probably because the proof that every matrix with trace zero is a commutator is difficult (although I'm not sure that this is the case).

Hope that clears things up a bit. If not, just ask!

share|improve this answer
    
Ah, I see. I seem to have misinterpreted the question. I remember reading a while ago that every traceless matrix is in fact a commutator, I was probably primed by that fact into thinking the question meant the subspace of commutators. Thank you for the quick clarification. –  EuYu Jan 1 '12 at 7:20
1  
Just something that my previous comment inspired. If it is true that every matrix with zero trace is the commutator of a pair of matrices, and that the set of traceless matrices form a subspace, how can it be true that the set of commutators do not? –  EuYu Jan 1 '12 at 7:28
    
Dear Alex: I don't understand your answer. Thanks in advance for your help. –  Pierre-Yves Gaillard Jan 1 '12 at 7:33
    
@EuYu: I looked it up and you are in fact right. I guess the question simply didn't want you to have to prove that. I am editing my answer appropriately. –  Alex Becker Jan 1 '12 at 7:39
    
@Pierre-YvesGaillard I edited it to fix a major blunder. Do you still not understand, and if so what are you not understanding? –  Alex Becker Jan 1 '12 at 7:44

If the ground field $\mathbb{F}$ is $\mathbb{R}$ or $\mathbb{C}$, the following gives an elementary proof. Clearly every commutator has zero trace, so it suffices to show that every real or complex matrix with zero trace is a commutator. First, every traceless matrix is $\mathbb{F}$-similar to a matrix with zero diagonal (we will prove this claim later). So, WLOG we may assume that $C$ has a zero diagonal. Take $A$ as an arbitrary diagonal matrix $\mathrm{diag}(a_1,\ldots,a_n)$ with real and distinct diagonal entries. The equation $C=AB-BA$ then boils down to $(a_i-a_j)b_{ij}=c_{ij}$, which is solvable as $b_{ij}=c_{ij}/(a_i-a_j)$. QED

We now prove our claim that any traceless matrix $C$ is $\mathbb{F}$-similar to a matrix with zero diagonal when $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$. In the real case, as $C$ is traceless, if it has some nonzero diagonal entries, some two of them must have different signs. WLOG assume that they are $c_{11}$ and $c_{22}$. Note that $$ \begin{pmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{pmatrix} \begin{pmatrix}c_{11}&c_{12}\\c_{21}&c_{22}\end{pmatrix} \begin{pmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{pmatrix} =\begin{pmatrix}c_{11}\cos^2\theta-(c_{12}+c_{21})\sin\theta\cos\theta+c_{22}\sin^2\theta&\ \ast\\ \ast&\ \ast\end{pmatrix}. $$ Since $c_{11}$ and $c_{22}$ have different signs, $c_{11}\cos^2\theta-(c_{12}+c_{21})\sin\theta\cos\theta+c_{22}\sin^2\theta=0$ is always solvable over $\mathbb{R}$. We now turn to the complex case. By unitary triangulation, we may assume that $C$ is upper triangular. As $C$ is traceless, if it has some nonzero diagonal entries, some two of them must be distinct. Again, assume that they are $c_{11}$ and $c_{22}$. Perform diagonalization on the leading 2-by-2 principal block of $C$, we may further reduce it to a diagonal 2-by-2 block. Now we have $$ \frac{1}{1+z^2}\begin{pmatrix}1&-z\\ z&1\end{pmatrix} \begin{pmatrix}c_{11}&0\\0&c_{22}\end{pmatrix} \begin{pmatrix}1&z\\-z&1\end{pmatrix} =\frac{1}{1+z^2}\begin{pmatrix}c_{11}+c_{22}z^2&\ \ast\\ \ast&\ \ast\end{pmatrix}. $$ As $c_{11}$ and $c_{22}$ are nonzero and distinct, $c_{11}+c_{22}z^2=0$ is always solvable for some $z\in\mathbb{C}$ with $z^2\not=-1$. Therefore, in both the real and complex cases, the $(1,1)$-th entry of $C$ can be made zero via a certain similarity transform. Continue in this manner recursively for the trailing principal submatrices of $C$, we obtain a matrix with zero diagonal.

Afternote: The above proof has made use of many properties of matrices over $\mathbb{R}$ and $\mathbb{C}$, so I am not sure whether the idea of proof is applicable when the ground field is different. If not, clearly some people will find the proof dissatisfactory because it does not reveal the true reason why the set of commutators is a matrix subspace. This is reminiscent of the Cayley-Hamilton Theorem for real matrices, for which we can embed the ground field into $\mathbb{C}$ and prove the theorem easily for those unitarily diagonalizable matrices first, then use a continuity argument to finish the proof. Algebraists usually regard this proof as dissatisfactory, but those who mostly work on $\mathbb{R}$ and $\mathbb{C}$ may take a different view. At any rate, the following reference contains a relatively short proof (which fills two and a half pages) of the statement that every traceless matrix over a general field is a commutator:

A.A. Albert and Benjamin Muckenhoupt (1957), "On matrices of trace zeros". Michigan Math. J., 4(1):1-3.

share|improve this answer
    
"Perform unitary diagonalization on the leading 2-by-2 principal block of C" -- this one I do not see, that it is always possible? $\pmatrix{c_{11} & c_{12} \\ 0 & c_{22}} \sim \pmatrix{c_{11} & 0 \\ 0 &c_{22}}$ unitarily? –  adam W Dec 29 '12 at 14:52
    
@adamW It's a typo. There's no "unitary" and the diagonalization is a result of an ordinary similarity transform. –  user1551 Dec 29 '12 at 15:26
    
Nice result, and nice reference. Thanks! –  Andres Caicedo Jun 6 '13 at 3:43

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.