Eigenvalues of $UV^T$ in SVD decomposition

eigenvalues-eigenvectorsmatricessvd

After performing a singular value decomposition (SVD) of a real square matrix $A$,

$$A=USV^T$$

  1. How to prove that the absolute value of all eigenvalues of $UV^T$ are one?

  2. Is there any relation between the eigenvalues of $UV^T$ and those of $A$?

Best Answer

Let $Q=UV^T$, then $Q$ is orthogonal, since $QQ^T=UV^TVU^T=I$.

Now, if $\lambda$ is an eigenvalue of $Q$ for the eigenvector $v$, then $Qv=\lambda v$, hence $v^Hv=v^HQ^HQv=\bar\lambda\lambda v^Hv$ implies that $|\lambda|=1$. Note that $Q^T=Q^H$ since $Q$ is real.

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