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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 $A$?

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  • $\begingroup$ What do you mean by the absolute value of ... are one? $\endgroup$ – mathcounterexamples.net Oct 18 '19 at 11:34
  • $\begingroup$ @mathcounterexamples.net What is meant is "The absolute value of each eigenvalue equals 1". Of course $A$ must be square, otherwise te SVD is still defined, but not the eigenvalues of $UV^T$. $\endgroup$ – StayHomeSaveLives Oct 18 '19 at 11:41
  • $\begingroup$ If $\lambda=a+ib$, then the absolute value is $\rm{abs(\lambda)}=\sqrt{a^2+b^2}$ $\endgroup$ – AJHC Oct 18 '19 at 11:46
  • $\begingroup$ A late hint could be that all (non-real) complex roots to a real coefficiented polynomial equation come in conjugate pairs. $\endgroup$ – mathreadler Oct 18 '19 at 12:18
  • $\begingroup$ Eigenvalues of $UV^T$ are one in absolute value and eigenvalues of $A$ can be arbitrary. $\endgroup$ – Algebraic Pavel Oct 18 '19 at 12:55
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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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  • $\begingroup$ And regarding my second question, there is not a direct relation, or there is? $\endgroup$ – AJHC Oct 18 '19 at 11:51
  • $\begingroup$ @AJHC None that I know of - which does not mean there is none. However, bear in mind that all matrices that have identical $U$ and $V$ but different singular values will have the same value of $UV^T$. This implies that matrices with widely different eigenvalues have the same $UV^T$. $\endgroup$ – StayHomeSaveLives Oct 18 '19 at 18:41

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