So I'm trying to prove, that for $V\in\mathbb{B}(H, K)$ where $H, K$ are Hilbert's spaces that $V$ is a partial isometry if and only if $V^*$ is a partial isometry, where $V^*$ denotes Hermitian adjoint operator of $V$.
So, I know that $\|Vx\|= \|x\|, \forall x\in \operatorname{Ker} V^{\perp}$ and also that $\operatorname{Im}V^*\subseteq\overline{\operatorname{Im}V^*}=\operatorname{Ker}V^{\perp}$. So if I take $y\in K$ then it is $\|VV^*y\| = \|V^*y\|$ and here I'm stuck...
I also know that if i have implication $V$ is partial isometry $\Longrightarrow$ $V^*$ is partial isometry then other implication is straightforward cause of the $(V^*)^*=V$.
I'd appreciate some help.
To show that $T^*$ is a partial isometry, we need to show $T^*$ is an isometry from $\text{Ker}(T^*)^{\bot}$ to $\text{Im}(T^*)$. Now $\text{Im}(T)$ is dense in $\text{Ker}(T^*)^{\bot}$, thus it suffices to show that $T^*$ is an isometry on $\text{Im}(T)$. Now we have $$\text{Ker}(T)^{\bot}\xrightarrow{T}\text{Im}(T)\xrightarrow{T^*}\text{Im}(T^*) \subset\text{Ker}(T)^{\bot}.$$ Let $x,y\in \text{Ker}(T)^{\bot}$, we have $\langle Tx, Ty\rangle=\langle x,y\rangle$, which implies by duality $\langle (T^*T-I)x,y\rangle=0$. Thus $T^*T-I$ equals zero as an operator on $\text{Ker}(T)^{\bot}$. Thus $\|T^*Tx\| =\|x\|=\|Tx\|$ for all $x\in \text{Ker}(T)^{\bot}$, which implies $T^*$ is an isometry on $\text{Im}(T)$.