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Let $A,B\in M_n(\mathbb{R})$ be real (or complex) square matrices. Generally speaking,

$${\rm trace}(AB)\neq {\rm trace}(A){\rm trace}(B)$$

There are a lot of 'easy' examples where this doesn't hold. What is not obvious to me is whether there is any chance of it working for something other than $A=B=0$, the zero matrix. Even if you work out the details for $n=2$ the algebraic condition is not very insightful.

Question: What are sufficient or necessary conditions on $A$ and $B$ under which the trace is multiplicative, i.e., ${\rm trace}(AB)= {\rm trace}(A){\rm trace}(B)$?

Edit: This post from 2014 asked whether the trace is multiplicative or not. I am well aware that it is not, but a distinct question is whether there are conditions on $A$ and $B$ under which the trace becomes multiplicative.

Amir Sagiv
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    Why would you need this property? It doesn't hold for even the simplest cases, such as diagonal matrices. – Dmitry Aug 08 '20 at 15:48
  • Oh I agree it does not seem plausible, but maybe there is some structure with which it does work, or rather there's a proof that it's simply always wrong. – Amir Sagiv Aug 08 '20 at 15:49
  • @DietrichBurde Not quite - I'm well aware that for many, many cases ${\rm tr}(AB)\neq {\rm tr}(A){\rm tr}(B)$. My hope is that there is some condition, restrictive or not, underwhich the trace is multiplicative. Or conversely, that if the trace is multiplicative, than it implies something very strong – Amir Sagiv Aug 08 '20 at 17:15
  • @DietrichBurde: The question already mentions zero matrices. A slightly more general case would be matrices where the $(1,1)$-entry is the only non-zero one. – celtschk Aug 08 '20 at 17:54
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    @celtschk And also this "result" doesn't seem to be interesting at all, at least not much more than the zero solution. Of course other people might see this differently. – Dietrich Burde Aug 08 '20 at 17:55
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    Somewhat related: https://math.stackexchange.com/q/1267700/339790 and https://math.stackexchange.com/q/3418931/339790 – Rodrigo de Azevedo Aug 09 '20 at 08:38

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Not really an answer, but too long for a comment.

Consider just diagonal matrices. Let \begin{align} A = \begin{bmatrix} a & 0\\ 0 & a \end{bmatrix} \ \ \text{ and } \ \ B = \begin{bmatrix} b & 0\\ 0 & b \end{bmatrix}, \end{align} then \begin{align} \operatorname{tr}(AB) = 2ab = 4ab= \operatorname{tr}(A)\operatorname{tr}(B). \end{align} This means either $a$ or $b$ must be zero. So, whatever conditions you have must rule out scalar multiples of the identity.

In general, let (you can do this by diagonalizing one of the matrices) \begin{align} A = \begin{bmatrix} a & b\\ c & d \end{bmatrix} \ \ \text{ and } \ \ B = \begin{bmatrix} \lambda_1 & 0\\ 0 & \lambda_2 \end{bmatrix}. \end{align} Then \begin{align} \operatorname{tr}(AB)= \lambda_1 a+\lambda_2 d = (a+d)(\lambda_1+\lambda_2)=\operatorname{tr}(A)\operatorname{tr}(B) \end{align} if and only if $\lambda_1 d+\lambda_2 a = 0$. This means $[\lambda_1, \lambda_2]^T\perp [d, a]^T = M [a, d]^T$ where $M = \begin{bmatrix} 0 & 1\\ 1 & 0\end{bmatrix}$ .

Jacky Chong
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