Singular Values for $A$ and $A^2$

Question

$\forall A \in R^{n \times n}$, 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?

Answer 1

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 $0.5510229$

Answer 2

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 $A$.