# when is $AB = B^TA$ true?

In what particular situation would the following be true?

$$AB = B^TA$$

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

I also know that $$BB = B$$.

• Note that you question amounts to "when is $AB$ symmetrical?" – Arnaud Mortier Aug 17 at 17:21
• 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 $1$. – Ted Shifrin Aug 17 at 17:21
• @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. – Arnaud Mortier Aug 17 at 17:22
• @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. – 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.

• That was my comment. Yet you can try to do more. – 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 $$2$$).