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Suppose I have a general $n\times n$ real matrix $A$. And suppose that $A$ has an SVD of the form $A=U^T S V$ with S of the form $I_m \oplus D$ where $I_m$ is the identity $m\times m$ matrix and $D$ is a matrix of size $n-m \times n-m$.

This means that $A$ has $m$ singular values equal to 1. Would this suffice to conclude that $A$ has $m$ eigenvalues of modulus 1? Why? Why not?

glS
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user556621
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1 Answers1

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In general the eigenvalues have no direct relation to the singular values. The only thing you can really be sure of is that the eigenvalues, in magnitude, lie in the interval $[\sigma_n,\sigma_1]$. Also each singular value of zero is in fact an eigenvalue (with the corresponding right singular vector as an eigenvector).

The exception is when $A$ is unitarily diagonalizable, which is equivalent to being normal. Then the left singular vectors and the right singular vectors basically coincide (differing by a complex sign at most), and are eigenvectors. In this case the singular values are just the moduli of the eigenvalues.

Ian
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  • Hi, was wondering if you have a reference for the statement that the absolute eigenvalues lie in the interval given by the smallest and largest singular values. Thanks. – Dylan Dijk Aug 23 '23 at 17:01
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    @DylanDijk I don't have a reference, but I can give a quick proof. On the high end, $\sigma_1$ is (either by definition or essentially by definition) $\sup_{| x |=1} | Ax |$. Any eigenvalue's magnitude is at most equal to this, since if $x$ is a corresponding unit eigenvector then $|\lambda|=| Ax |$. On the low end, either $\sigma_n=0$ in which case the lower bound is trivial, or it isn't. If it isn't, then $A$ (assumed square to begin with) is invertible. – Ian Aug 23 '23 at 17:08
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    @DylanDijk (Cont.) So if $\tau_i$ are the singular values of $A^{-1}$ and $\eta_i$ are its eigenvalues, then the previous argument tells you $\tau_1 \geq |\eta_i|$. Now just express $\tau_1$ and $\eta_i$ in terms of $\sigma_i$ and $\lambda_i$ to get the lower bound. – Ian Aug 23 '23 at 17:08