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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.