Eigenvalues and singular values of $3 \times 3$

Question

I'm asked to provide an example or disprove the existence of a $3\times 3$ matrix whose eigenvalues are all $0$ but has singular values $1$, $1/2$, $0$. My initial instinct was to consider the possible Jordan canonical forms of such a $3\times 3$ but I'm unsure if this is the best approach.

Answer 1

Consider $$T = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & \frac12 \\ 0 & 0 & 0\end{bmatrix}$$

We have $T^*T = \operatorname{diag}\left(0, 1, \frac14\right)$ so the singular values of $T$ are $\left\{0, 1, \frac12\right\}$.

---

Also note that looking at the Jordan form is not sufficient because matrices in general are not unitarily similar to their Jordan forms. Meaning if a matrix $T$ has Jordan form $J$, then $T^*T$ may not have the same eigenvalues as $J^*J$.

What you could do is take a look at Schur decomposition to conclude that every matrix is unitarily similar to an upper-triangular matrix. Then you could have taken $$T = \begin{bmatrix} 0 & a & b \\ 0 & 0 & c \\ 0 & 0 & 0\end{bmatrix}$$

and calculated its singular values.