# If $d$ is a singular value of an operator $T$, then is $d^2$ a singular value of $T^2$?

I'm trying to prove/disprove a homework problem that is the title question. I'm not looking for an explicit answer, just some direction. So, I've been reading Axler's book Linear Algebra Done Right and from what I understand,

every operator on $V$ has a diagonal matrix with respect to some orthonormal bases of $V$, provided that we are permitted to use two different bases rather than a single basis as customary when working with operators.

So that if we let $( e_1, ..., e_n )$ and $( f_1, ..., f_n )$ be orthonormal bases of $V$ then the matrix of $T$, an operator on $V$, is

$$M(T,(e_1,...,e_n),(f_1,...,f_n)) = \begin{bmatrix} d_{1} & 0 & \cdots & 0 \\ 0 & d_{2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & d_{n} \end{bmatrix}$$

So, my thinking is that the matrix of $T^2$ is just the matrix above composed with itself and so would be

$\begin{bmatrix} d^2_{1} & 0 & \cdots & 0 \\ 0 & d^2_{2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & d^2_{n} \end{bmatrix}$

Am I on the right track or am I missing something?

• I've no idea what the answer is since the above seems to be a rather unique, specialized thing, but could you give the exact reference in Axler's book? Commented Mar 7, 2014 at 14:11
• Do you mean the exact page number of where I took the quote? If so, it's on page 157. The chapter title is "Operators on Inner-Product spaces" (chapter 7). Commented Mar 7, 2014 at 19:01

## 1 Answer

Think of singular values this way: $T$ will rotate the unit ball, then stretch it to a coordinate-aligned ellipsoid with semiaxes $d_1,\dots,d_n$, then rotate that ellipsoid.

Imagine doing that twice. Will the stretch factors be $d_1^2,\dots,d_n^2$? Probably not, because we have rotations between stretching maps, which will spoil everything.

Once this is realized, a counterexample comes naturally: let $T$ be something that deforms and then rotates. For example: $$T=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$ has $1$ as its singular value, but $T^2$ is the zero matrix.