What conditions could make the commutator $[M^,M] = M^M - MM^*$ full-ranked? $M$ is not normal

35 Views Asked by Bumbble Comm At 28 Mar 2026 - 6:59

Let $M$ be complex, full-ranked, and not normal i.e. $M\ne M^*$.

Is $M^*M - MM^*$ also full-ranked?

Edit:

If there is any, what condition(s) could make the commutator $[M^*,M] = M^*M - MM^*$ full-ranked? $M$ is not normal.

If it helps, I know that $M^*M$ is full-ranked which implies that $MM^*$ is also full-ranked.