In what particular situation would the following be true?

$AB = B^TA$

where $A$ is symmetrical, $B$ is not.

I also know that $BB = B$.

  • 1
    $\begingroup$ Note that you question amounts to "when is $AB$ symmetrical?" $\endgroup$ – Arnaud Mortier Aug 17 at 17:21
  • $\begingroup$ I don't think the condition $B^2=B$ makes much difference. $B$ and $B^\top$ are always similar, and this equation is the change of basis condition that says $B = A^{-1}B^\top A$, so $A$ gives the similarity. If it helps, $B^2=B$ tells you that $B$ can be diagonalized and has eigenvalues $0$ and $1$. $\endgroup$ – Ted Shifrin Aug 17 at 17:21
  • $\begingroup$ @TedShifrin It is not given or required that $A$ be invertible here, however. As for $B$, another way to say it is that it is a projection. $\endgroup$ – Arnaud Mortier Aug 17 at 17:22
  • $\begingroup$ @ArnaudMortier: Yes, I'm fully aware of that (a not-necessarily orthogonal) projection. And, yes, of course, the equation holds with $A=O$, for example, but that doesn't seem so interesting. In the invertible case, that the change-of-basis matrix $A$ is symmetric is interesting, I guess. $\endgroup$ – Ted Shifrin Aug 17 at 17:28

We know that $$(AB)^T=B^TA^T=B^TA=AB$$therefore the symmetry of $AB$ is an equivalent condition. No more general condition can be implied.

  • $\begingroup$ That was my comment. Yet you can try to do more. $\endgroup$ – Arnaud Mortier Aug 17 at 18:28

Let us do it in dimension $2\times 2$.

$A=\pmatrix{a &b\\ b&c}$ and $B=\pmatrix{s&p\\q&r}$.

The conditions on $B$ are:

It's not symmetric

  • $p\neq q$

It's a projection

  • $p(s+r)=p\qquad$ and $\qquad q(s+r)=q$
  • $s^2+pq=s\qquad$ and $\qquad r^2+pq=r$

$AB$ is symmetric

  • $ap+br=sb+qc$

First deduction: $r+s= 1$. Indeed, if not then $p=q=0$ but we know that $p\neq q$.

Second deduction: $r$ and $s$ are the two solutions to the equation $x^2-x+pq=0$. That is $$r,s=\frac{1\pm \sqrt{1-4pq}}{2}$$ and in particular, we must have $1-4pq\geq 0$.

Third deduction: $$ap+b\sqrt{1-4pq}=qc$$

From this we can derive a method to produce all possible matrices $B$:

  • Pick any value for $q$,
  • Solve (in $p$) the quadratic equation $(ap-qc)^2=b^2(1-4pq)$, keep only those such that $p\neq q$, such that $1-4pq\geq 0$ (if $b\neq 0$ this comes for free from the quadratic equation) and then keep only those which also satisfy $ap+b\sqrt{1-4pq}=qc$.
  • For each of the solutions found for $p$, set $r=\frac{1+ \sqrt{1-4pq}}{2}$ and $s=\frac{1- \sqrt{1-4pq}}{2}$: your matrix $B$ is fully defined and has all criteria.

Example: pick $q=0$. Assuming $a\neq 0$, the relevant solution to the quadratic equation is $p=-\frac ba$. Assuming $b\neq 0$ we have $p\neq q$. Here, $1-4pq\geq 0$ comes for free. $r=1$ and $s=0$. This gives $$B=\pmatrix{0 & -\frac ba\\ 0 & 1}$$ You can check that $AB$ is symmetric as required: $$\pmatrix{a &b\\ b&c}\pmatrix{0 & -\frac ba\\ 0 & 1}=\pmatrix{0 &0\\ 0&c-\frac{b^2}{a}}$$ and that the other requirements are met as well.

Conclusion: In general, there are a lot of such matrices (potentially several $1$-parameter families, which can be described explicitly at least in dimension $2$).


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