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Suppose $C$ is an $n\times n$ matrix over complex numbers, with trace $0$.

Are there always $n\times n$ matrices $A,B$ such that $AB - BA = C$?

(Inspired by a recent question which asked for a trace free proof of non-existence of solutions for $C=I$).

Martin Sleziak
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me.dorgan
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3 Answers3

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Yes, this is always true. See a proof, for example, over here. The statement is proven by induction on $n$.

The key to the proof presented in the link is the following proposition:

Lemma 2: if $S \neq \lambda I$ for any scalar $\lambda \neq 0$, then $S$ is similar to a matrix with a $0$ in the $(1,1)$ entry.

There is a similarly useful extension of this statement in Horn and Johnson which says that every matrix is similar to some matrix whose diagonal entries are identical.

Ben Grossmann
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    @Omnomnonmonm, i have a question about the proof of lemma2 in the paper your linked. i understand the claim that $w^Tv = 1$ and $w^TSv = 0$ cannot hold. author claims that this in turn implies $w^TS = \mu w^T$ for some $\mu \neq 0.$ that i don't see why. do you understand why? – abel Dec 04 '14 at 07:57
  • @abel here's one justification: both $w^T$ and $w^TS$ are linear maps $\Bbb R^n \to \Bbb R$. The fact that for all $v$: $w^TSv = 0 \implies w^Tv = 0$ (a corollary to the statement shown) indicates that $\ker w^T \subseteq \ker w^TS$. This implies that $w^TS = \mu w^T$ for some $\mu \neq 0$. – Ben Grossmann Dec 04 '14 at 13:48
  • @Omnomnonmonm, thanks. he really uses if $w^Tv \neq 0,$ then $w^tSv \neq 0$ and its contra positive statement $w^TSv = 0,$ then $w^Tv = 0$ to show $w^TS = \mu w$ for some $\mu \neq 0.$ is this what he is doing? – abel Dec 05 '14 at 04:25
  • @abel as far as I can tell, yes: that's what he's doing. – Ben Grossmann Dec 05 '14 at 05:03
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this is not an answer but a longer comment. let me try the case $n = 2.$ take the matrices $\pmatrix{a_1 & a_2 \cr a_3 & a4}, B = \pmatrix{b_1 & b_2 \cr b_3 & b_4}, \mbox{ and} \pmatrix{c_1 & c_2\cr c_3 & -c_1}.$ writing out $AB - BA = C$ as a string of linear equations i get an over determined system $$\pmatrix{0 & 0 & a_2 & -a_3\cr -a_2 & a_1 - a_4 & 0 & a_2\cr a_3 & 0 & a_4 - a_1 & -a_3} \pmatrix{b_1 \cr b_2\cr b_3 \cr b_4} = \pmatrix{c_1 \cr c_2 \cr c_3} $$

of course, one has to worry about inconsistency.

abel
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  • No, the first line of the $3 \times 4$ matrix is erroneous ; this matrix should be $$\pmatrix{0 & -a_3 & a_2 & 0 \cr -a_2 & a_1 - a_4 & 0 & a_2\cr a_3 & 0 & a_4 - a_1 & -a_3} $$ and the problem is that this matrix has rank 2 at most... Thus beeing given a RHS, it is far from being sure that this RHS is in the range space of this matrix – Jean Marie Jan 09 '19 at 22:27
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In fact, there are many solutions in $(A,B)\in \mathbb{C}^{2n^2}$. Let $f:(A,B)\rightarrow AB-BA\in \{U;tr(U)=0\}\approx \mathbb{C}^{n^2-1}$. It can be shown that, for a generic $(A,B)$, $f$ is a submersion in a neighborhood of $(A,B)$ (that is $Df_{A,B}$ is surjective). Then, for a generic $C\in \mathbb{C}^{n^2}$, $f^{-1}(C)$ is an algebraic set of dimension $2n^2-(n^2-1)=n^2+1$ (degrees of freedom).

Beware 1. it is not true for particular $C$; for example, $f^{-1}(0_n)$ has dimension $n^2+n$.

Beware 2. Although $dim(f^{-1}(C))>n^2$, we cannot randomly choose $A$ (for example).

EDIT. Since $rank(Df_{A,B})\leq n^2-1$, the minimum of $dim(f^{-1}(C))$ is $n^2+1$ (generic case). Now, about the maximum, I think that it is $n^2+n$ but I am not sure...