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$\forall A \in R^{n \times n}$, if $\sigma$ is a singular value of $A$, then $\sigma^2$ is a singular value of $A^2$

Intuitively, this seems to be false, but how can I prove this?

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    $\begingroup$ Find a nonzero matrix $A$ that $A^2=0$. Compare the singular values. $\endgroup$
    – A.Γ.
    Dec 13, 2019 at 23:11

2 Answers 2

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As А.Г. commented, any nonzero matrix $A$ with $A^2=0$ will do for a counterexample. For instance, the singular values of $\small{\begin{bmatrix}0&1\\0&0\end{bmatrix}}$ are $0$ and $1$, but the singular values of the zero matrix are, of course, all zero.

As to why this might not be the case even for non-nilpotent matrices, consider the interpretation of the SVD as decomposing the transformation represented by the matrix into a rotation/reflection, then a nonuniform scaling, followed by another rotation/reflection. If none of the singular values is zero, we can interpret them as the semiaxis lengths of the ellipsoid that is the image of the unit sphere under this transformation. If you apply the same transformation to this ellipsoid, its axes generally won’t line up with the directions in which the scaling occurs. You’ll still end up with an ellipsoid, but its semiaxis lengths aren’t going to be the squares of those of the original ellipsoid. This isn’t a formal proof, but it should give you a feel as to why it’s going to be unusual for all of the singular values of $A^2$ to be the squares of the singular values of $A$.

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  • $\begingroup$ What do you mean by find any nonzero matrix $A$ s.t. $A^2=0$? Does this mean that $A^2$ is a zero matrix? $\endgroup$
    – Eisen
    Dec 14, 2019 at 1:50
  • $\begingroup$ @Eisen Just so. $\endgroup$
    – amd
    Dec 14, 2019 at 4:29
  • $\begingroup$ @amd I'd have a question: assuming $A$ is a $n \times n$ real, positive definite, symmetric matrix, of which we know its eigenvalues $\lambda = (\lambda_1 > \lambda_2 > ... > \lambda_n)$ and eigenvectors, can we relate the eigenvalues and vectors of $A^2$ knowing them for $A$ ? $\endgroup$
    – mEm
    Jan 16, 2023 at 9:35
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IT is enough to show a counter example. Consider a 2 by 2 matrix $$ A= \begin{pmatrix} -1 & -6 \\ 0 & 2 \end{pmatrix} $$

The singular values of A are $6.3954832$ and $0.3127207$. On the other hand $$ A^2= \begin{pmatrix} 1 & -6 \\ 0 & 4 \end{pmatrix} $$ has singular values $7.2592268$ and $0.5510229$

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  • $\begingroup$ Not sure one has to go so far. The counterexample in @amd answer looks way too simpler. $\endgroup$
    – A.Γ.
    Dec 14, 2019 at 21:52

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