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$A$ and $B$ are given $n\times n$ real symmetric matrices. If we have

$$\det(I+xA)\det(I+xB) = \det(I+xA+xB)$$

for all $x\in \mathbb{R}$, then prove $AB=0$.

I've tried expanding the determinant to $$\det(I+xA+xB+x^2AB) = \det(I+xA+xB), \forall x\in \mathbb{R}$$ Letting $C=A+B$ and $D=AB$, $C$ is symmetric so can be diagonalized, but $D$ may not. I also tried letting $x\to \infty$ and get $\det(AB)=0$. But I still cannot get through it. Am I along the wrong way?

Cameron L. Williams
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Enthalpia
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    On AoPS: https://artofproblemsolving.com/community/c7h1915990p13130527. I did not verify that solution. – Martin R Jun 26 '25 at 05:54
  • @MartinR Thanks! But that problem is not exactly the same. The condition here is restricted to $x=y$ case, instead of $\forall x,y$. – Enthalpia Jun 26 '25 at 06:22
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    It looks to me like differentiating wrt x should give an elegant solution but I can't quite get it to work. – quarague Jun 26 '25 at 09:06
  • @BrianMoehring I was thinking to same thing. But in order to use some degree/polynomial argument you have to argue that the coefficient in front of the $x^2$ is zero implies that $AB=0$, which is basically the whole point. – F. Conrad Jun 26 '25 at 16:31
  • This question is almost an exact duplicate of https://math.stackexchange.com/q/4657419 , but it was closed as a duplicate of https://math.stackexchange.com/q/87982 because the latter question had received more answers. – user1551 Jun 26 '25 at 17:41
  • @F.Conrad. For whatever reason, I had completely misread the problem to say they were trying to show $AB$ was singular. In any case, I now agree my previous comment wasn't useful. – Brian Moehring Jun 26 '25 at 17:51
  • @BrianMoehring No worries, I was having the EXACT same thought as you did and just wanted to share it. – F. Conrad Jun 26 '25 at 19:02

1 Answers1

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We need to make use of the condition that $A$ and $B$ are real symmetric matrices, otherwise the assertion that $AB=0$ may be false. For example, let $A=I$ and $N$ be the nilpotent matrix $\pmatrix{0&1\\ 0&0}$. The determinantal equality is then satisfied when $B=N$ (so that $A$ and $B$ are real but $B$ is not symmetric) or when $B=\pmatrix{2&1\\ 2i&1}N\pmatrix{2&1\\ 2i&1}^{-1}=\pmatrix{1-i&1+i\\ 1+i&i-1}$ (so that $A$ and $B$ are symmetric but $B$ is not real), but $AB=B\ne0$.

Anyway, suppose $A$ and $B$ are real symmetric. The determinantal condition implies that $\det(yI-A)\det(yI-B)=y^n\det(yI-A-B)$ for every real number $y$. Hence $\pmatrix{A\\ &B}$ and $\pmatrix{A+B\\ &0}$ have the same characteristic polynomial, and they must be similar because they are real symmetric. In particular, $$ d+s:=\operatorname{rank}(A)+\operatorname{rank}(B)=\operatorname{rank}(A+B)\le n. $$ Let $A=UDU^T$ and $B=VSV^T$ where each of $U\in M_{n,d}(\mathbb R)$ and $V\in M_{n,s}(\mathbb R)$ has orthonormal columns and $D\in M_{d,d}(\mathbb R),\,S\in M_{s,s}(\mathbb R)$ are invertible diagonal matrices containing the nonzero eigenvalues of $A$ and $B$ respectively. From the similarity between $$ \pmatrix{A\\ &B}\quad\text{and}\quad\pmatrix{A+B\\ &0}, $$ we infer the similarity between $$ \pmatrix{D\\ &S\\ &&0_{(n-d-s)\times(n-d-s)}} \quad\text{and}\quad A+B=\pmatrix{U&V}\pmatrix{D\\ &S}\pmatrix{U^T\\ V^T}. $$ In turn, $$ \pmatrix{D\\ &S} \quad\text{and}\quad \pmatrix{U^T\\ V^T}\pmatrix{U&V}\pmatrix{D\\ &S} $$ have the same characteristic polynomial. Therefore $$ \det\pmatrix{I_d&U^TV\\ V^TU&I_s} =\det\left(\pmatrix{U^T\\ V^T}\pmatrix{U&V}\right) =1 $$ and by Hadamard’s inequality for positive definite matrices (i.e., $\det P\le\prod_i p_{ii}$ whenever $P$ is PD, and equality holds iff $P$ is diagonal), we must have $U^TV=0$. Hence $AB=UDU^TVSV^T=0$.

user1551
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  • isn't this a duplicate of https://math.stackexchange.com/questions/87982/old-amm-problem/ or https://math.stackexchange.com/questions/4657419/let-a-b-in-m-n-times-n-mathbbr-be-real-symmetric-matrices-prove-that-a/? – user8675309 Jun 26 '25 at 17:34
  • @user8675309 Well, I guess I am growing old and have become absent-minded. – user1551 Jun 26 '25 at 17:38