Prove $ST = 0 \implies ST^* = 0$.

Question

EDIT - checking if it is better now

So I have some proof and I'm not sure on one of the transitions.

---

In this same assignment we had to prove the following:

Let V a Hermitian vector space and $T:V \to V$ is normal in $V$. Prove $T,T^*$ have the same kernel and Image.

Using the above theorem I solved the current question:

Let V a Hermitian vector space $T:V \to V$ is normal in $V$ and $S:V \to V$ is linear. Prove $ST=0 \implies ST^* = 0$.

---

My proof is this:

Suppose $ST=0$. Since $T$ is normal in $V$ and $V$ is Hermitian:

we have Ker$T$ = Ker$T^*$ and Im$T$=Im$T^*$.

Let some $v \in V$ then from the datum $ST(v)=0$. We have 2 options:

1. $T(v) = 0 \implies S(T(v)) = S(0) = 0$
2. $T(v) = u \neq 0 \in V,\;S(u) = 0$

For case #1 since we have Ker$T$=Ker$T^*$ then: $T(v) = 0 \iff T^*(v) = 0$ we can derive $ST = 0 \implies ST^* = 0$

For case #2 we have $ST = 0 \implies$ Im$T \subseteq$ Ker$S$. Using Im$T$ = Im$T^*$ we can deduce Im$T^* \subseteq$ Ker$S$ therfore implying $ST^*=0$

All in all we have $ST=0 \implies ST^* = 0$

Answer 1

Your proof is correct, and this statement is not true in general for non-normal operators. Let $\{e_0, e_1, \dots\}$ be a basis for your Hilbert space and consider the shift operator $T(e_i) = e_{i+1}$ and $S(e_i) = 0\ \forall\ i>0$ then $ST=0$, I'll leave it to you to figure out $T^*$ and see that the composition is not zero.