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I'm trying to find an example to show that the product of two non-zero symmetric matrices can be anti-symmetric.

I've proven that this is impossible for 2x2 matrices.

For 3x3 matrices, I've formulated a set of linear equations in 12 variables and used MATLAB to try and find a solution, to no avail.

So, is this possible, and if so, what is the best method to use to formulate an example? If not, what is the best way to prove that it is impossible for matrices of size n (n arbitrary natural number)?

With very many thanks,


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What wrong with $\operatorname{diag}(1,0,\ldots,0)$ and $\operatorname{diag}(0,1,0,\ldots,0)$? – Davide Giraudo May 6 '12 at 14:48
I don't understand the meaning of diag... please could you explain? – Froskoy May 6 '12 at 15:01
$\operatorname{diag}(a_1,\ldots,a_n)$ a diagonal matrix (hence symmetric) such that the first diagonal entry is $a_1$, the second $a_2$, the third $a_3$, etc... – Davide Giraudo May 6 '12 at 15:04
Doesn't that just give the zero matrix? Is there are solution that is not equal to the zero matrix? I guess the zero matrix is antisymmetric though. – Froskoy May 6 '12 at 15:07
By $A$ is anti-symmetric, do you mean that the transpose of $A$ is $-A$? – Davide Giraudo May 6 '12 at 15:11
up vote 3 down vote accepted

Of course, it's not possible in dimension $1$. If the dimension $d$ is greater than $2$, then let $A=(a_{lr})_{1\leq l,r\leq d}$ and $B=(b_{lr})_{1\leq l,r\leq d}$, with $a_{1,1}=1$, $b_{2,2}=0$ and all the other entries are $0$. Then $A$ and $B$ are non-zero, symmetric and $AB=0$, which is skew-symmetric.

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This is not an answer, just an observation. Matrices are all real.

You can show that if $A$ is symmetric definite positive and $B$ is any symmetric matrix, then $AB$ is diagonalizable (over $\mathbb{R}$) and hence can only be skew-symmetric when $B=0$, because the only skew-symmetric matrix which is diagonalisable (over $\mathbb{R}$) is the $0$ matrix. I wonder what happens if $A$ is assumed invertible but not of definite sign...

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Say you have two symmetric matrices, $A$ and $B$; In order for their product to be anti-symmetric, i.e, for $(AB)=-(AB)^T$, $AB=-BA$, i.e. they anti-commute.

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