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Let $A,B$ be $n\times n$ real nonsingular matrices ($n$ is even). $B$ is skew-symmetric, namely $B^T=-B$. Consider the quadratic eigenvalue problem $$ (A\lambda^2+B\lambda-A^T)\mathbf{x} =0 $$ with eigenvalue $\lambda$ and right eigenvector $\mathbf{x}$.

It is obvious that we have $2n$ eigenvalues and right eigenvectors. They appear in pairs as $\lambda$ and $1/\lambda^*$. Moreover, $\lambda$ and $\lambda^*$ also appear in pairs if they are complex. Assuming all of them have absolute value different from $1$, then $n$ of them will have absolute value less then $1$. Now we pick the corresponding right eigenvectors $\mathbf{x}_1,\dots,\mathbf{x}_n$, is it true that the rank of $(\mathbf{x}_1,\dots,\mathbf{x}_n)$ can only be $n$ or $n-1$?

I tried a lot of numerical experiments and found that it is indeed the case.

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