# Eigenvalues of $UV^T$ in SVD decomposition

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 $$A$$?

• What do you mean by the absolute value of ... are one? – mathcounterexamples.net Oct 18 '19 at 11:34
• @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 $UV^T$. – StayHomeSaveLives Oct 18 '19 at 11:41
• If $\lambda=a+ib$, then the absolute value is $\rm{abs(\lambda)}=\sqrt{a^2+b^2}$ – AJHC Oct 18 '19 at 11:46
• A late hint could be that all (non-real) complex roots to a real coefficiented polynomial equation come in conjugate pairs. – mathreadler Oct 18 '19 at 12:18
• Eigenvalues of $UV^T$ are one in absolute value and eigenvalues of $A$ can be arbitrary. – Algebraic Pavel Oct 18 '19 at 12:55

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.
• @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 $UV^T$. – StayHomeSaveLives Oct 18 '19 at 18:41