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Suppose $3\times 3$ complex symmetric matrix $$ A_{ab} = B_{ai}B_{bi}, $$ where $B$ is arbitrary complex $3\times 3$ matrix and $i$ is fixed.

How to show that it has two zero eigenvalues?

I've shown that $\text{det}A = 0$, which means that at least one of eigenvalues is zero, but this doesn't tell me how many zero eigenvalues it has. I've also shown that for the given row the other two are linearly dependent.

Edit: I mean, maybe there are other way to show that it has two zero eigenvalues except by calculating of its characteristic polynomial? For example, I then want to apply corresponding method for the matrix $$ A'_{ab} = B_{ai}B_{bi} + B_{aj}B_{bj} $$ to prove that is has only one zero eigenvalue, and its characteristic polynomial is expected to be large and tedious.

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  • $\begingroup$ I think you messed up the indices... $\endgroup$ – user394255 Jan 16 '17 at 11:07
  • $\begingroup$ @A.Molendijk : You're right. I meaned $A_{ab} = B_{ai}B_{bi}$ without summation over i. $\endgroup$ – John Taylor Jan 16 '17 at 11:09
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Let $X$ be the $i$-th column of $B$. Then, $A = X^tX$, which means $A$ is of rank $1$, thus, it needs to have at least two zero eigenvalues.

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