Matrices problem with determinant and rank

Question

Let us consider $A,B$ two $n \times n$ matrices with rational elements. Prove that if $(A+B)^2=AB$, then $\det(AB-BA)=0$.

My try: By the conclusion, I tried to prove that $\mathrm{rank}(AB-BA) \lt n$ and I thought about Sylvester's or Frobenius' inequalities. I can't go any further.

As it is stated the question is false. For a counterexample see here. — user26857, Jan 12 '21 at 22:25
I suppose the OP wanted to write $(A-B)^2=AB$. This question is solved here. — user26857, Jan 12 '21 at 22:40

score 1 · Accepted Answer · edited Jan 12 '21 at 21:31

1

Let $a = \frac{3+\sqrt{5}}{2}$ and $b = \frac{3-\sqrt{5}}{2}$, $ab = 1$.
Then, $\det(A-aB) = c+d\sqrt{5} $ and $\det(A-bB) = c-d\sqrt{5}$.
So $\det(A-aB)\det(A-bB)$ is a rational number.
Thus, by calculating $(A-aB)(A-bB)$ and then applying the determinant, we obtain that $\det(AB-BA) = 0$.

This problem is similar to the following problem:

Let $A$, $B$ be two $n\times n$ matrices with real elements. Prove that if $A^2+B^2=AB$, then $\det(AB−BA)=0$.
Here you take $\epsilon$ a non-real number, $\epsilon^3=1$, the root for the equation $x^2+x+1=0$.
Then you calculate $(A+\epsilon B)(A+\epsilon^2 B)$, apply the determinant, and then the conclusion is proved.

edited Jan 12 '21 at 21:31

user26857

52,094

answered Mar 21 '17 at 20:55

C_M

3,624

For the second claim you probably want $n$ not divisible by $3$. – user26857 Jan 12 '21 at 21:32
For $n=3$ you can find a counterexample here: https://math.stackexchange.com/questions/1768707 – user26857 Jan 12 '21 at 22:14
To conclude: the question is false and this answer as well. – user26857 Jan 13 '21 at 14:31

Matrices problem with determinant and rank

1 Answers1