Does the set of matrix commutators form a subspace?

Question

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?

Answer 1

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 zero". Michigan Math. J., 4(1):1-3.

Answer 2

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!

Answer 3

Let us denote by $M_n(\mathbb{R})$ the space of real square matrices of size $n$ and consider a diagonal matrix $D=[\lambda_i \delta_{i,j}]_{1\leqslant i,j\leqslant n}$ with $\lambda_1,\cdots,\lambda_n$ all distinct real numbers.

Then, the map $f:M_n(\mathbb{R})\to M_n(\mathbb{R}),M\mapsto DM-MD$ is linear and it is readily seen that its kernel is the set of all diagonal matrices.

Since the range of $f$ is a subset of the set Z of all matrices whose diagonal termes are all 0, the rank formula $\text{rank}(f)=n^2-\dim(\ker(f))=n^2-n$ proves that $\text{range}(f)=Z$.

Now, consider $C\in M_n(\mathbb{R})$ such that $\text{tr}(C)=0$. There exists $P\in GL_n(M_n(\mathbb{R}))$ and $N\in Z$ such that $C=P^{-1}NP$ (this can be shown by induction).

Moreover, there exists $M\in M_n(\mathbb{R})$ such that $N=DM-MD$.

Finally : $C=P^{-1}\left(DM-MD\right)P=AB-BA$ where $A=P^{-1}DP$ and $B=P^{-1}MP$.

Hence, every matrix with zero trace is a commutator. Since the converse is also true, the set of commutators appears to be the kernel of the trace (and, in particular, a subspace of $M_n(\mathbb{R})$).

Answer 4

(Edit: The earlier proof was flawed. Here is my revised version.)

This is a supplement of the proof of @user1551. In his proof, he proved the claim for $F=\mathbb{R}$ or $\mathbb{C}$. I generalized this claim to the field with characteristic $=0$ as the following.

Let $k$ be a field, $A\in k^{n\times n}$ is traceless. If $\mathrm{char}(k)=0$, $A$ is similar to some matrix with zero diagonal.

Proof If $\mathrm{char}(k)=0$.

We first show that for any $2\times 2$ matrix with two different diagonals, it is similar to some matrix with the first diagonal being zero. Take any matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ with $a\neq d$. We are finding some invertible matrix $\begin{pmatrix} e & f\\ g& h \end{pmatrix}$, such that: $$\begin{pmatrix} e & f\\ g& h \end{pmatrix} \cdot \begin{pmatrix} a & b \\ c & d \end{pmatrix}\cdot \begin{pmatrix} h & -f\\ -g& e \end{pmatrix} $$ has zero diagonal. This simplifies to (after a tedious calculation) $$ f(ch-dg)+e(ah-bg)=0 \quad (*) $$

(a) If $b\neq 0$, then we take $f=0$, and equation $(*)$ simplifies to $e(ah-bg)=0 $. Take $h=b,g=a$, then equation $(*)$ holds, and $\begin{pmatrix} e & f\\ g& h \end{pmatrix}=\begin{pmatrix} e & 0\\ a & b \end{pmatrix}$, we then take any non-zero $e$ to make it invertible.

(b) If $c\neq 0$, then we take $e=0$, and equation $(*)$ simplifies to $f(ch-dg)=0$. Take $g=c, h=d$, then equation $(*)$ holds, and $\begin{pmatrix} e & f\\ g& h \end{pmatrix}=\begin{pmatrix} 0 & f \\ c & d \end{pmatrix}$, we then take any non-zero $f$ to make it invertible.

(c)If $b=c=0$, then equation $(*)$ simplifies to $aeh-dfg=0$. Take any matrix $\begin{pmatrix} e & f\\ g& h \end{pmatrix}$ with $eh=d,fg=a$, then equation $(*)$ holds, and its determinant $d-a\neq 0$.

Now, take any traceless matrix $A\in k^{n\times n}$. Since $\operatorname{char}(k)= 0$, there exists $2$ different diagonals, say the first two. Then by the previous discussion, there exists $P\in GL_n(k)$, such that $$\begin{pmatrix} P^{-1} &0 \\ 0 & I_{n-2} \end{pmatrix} \cdot A \cdot \begin{pmatrix} P &0 \\ 0 & I_{n-2} \end{pmatrix} $$ is a matrix whose first diagonal being zero. We do this again for the remaining $(n-1)\times (n-1)$ principal submatrix, and so on. $\square$

(Note: If $\operatorname{char}(k)=p\neq 0$, the matrix $I_p$ is clearly a counter-example. )