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If there is a symmetric matrix, say

$$B = \left[\begin{array}{cc} 0 & A\\ A^T & 0 \end{array}\right]$$

where $A$ is a $m\times n$ submatrix with $m \geq n$.

Is it possible to express the eigenvalues and eigenvectors of $B$, in terms of singular values and singular vectors of $A$?

(Apologies for the bad formatting)

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1 Answer 1

I assume these are real matrices. Hint: The condition for $\pmatrix{u\cr v\cr}$ to be an eigenvector of $B$ for eigenvalue $\lambda$ is $A v = \lambda u$ and $A^T u = \lambda v$. What does that say about $A^T A v$?

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