For example, I have the following exercise: $\text{“Prove that if }N\text{ is normal and }AN=0\text{, then }AN^\star = 0”.$
I know that an $\textbf{Incorrect}$ way to prove this would be
$AN^\star = 0 \implies$ (Multiplying by $N$)
$AN^\star N = 0 \implies$ ($N$ is normal)
$ANN^\star = 0 \implies$ (Because $AN = 0$)
$0 = 0\;$
Q.E.D (The mistake here is that we assumed what we needed to prove)
But is the following proof correct?
First we'll prove a lemma: “If $AA^\star = 0$, Then $A = 0$.”
Let $V$ be a vector space.
$AA^\star = 0 \implies \forall x\in V : \langle AA^\star x, x \rangle = 0 \implies \langle A^\star x, A^\star x \rangle = 0 \implies ||A^\star x||^2 = 0 \implies A^\star x = 0 \implies A^\star = 0 \implies A = 0$
$AN^\star = 0 \iff$
$AN^\star (AN^\star)^\star = 0 \iff$(Conjugate properties)
$AN^\star NA^\star = 0 \iff (N$ is normal)
$ANN^\star A^\star = 0 \iff$ (Because $AN = 0)$
$0 = 0$
Which is true, and we used if-and-only-ifs and therefore $AN^\star = 0$.
Is the second proof correct?
Thanks!