Do partial isometries commute with their adjoints when applied to certain subspaces

Question

Let $T$ be a linear operator on a finite dimensional inner product space $V$. Let $W$ be a linear subspace of $V$.

For homework I would like to show that if $\forall x \in W, x' \in W^{\perp},\space \left\lVert Tx\right\rVert = \left\lVert x\right\rVert $ and $Tx'= 0$, then the same is true for $T^{*}$.

I know that $T^{*}Tx = x$ for $x \in W$, and I would like to say $\langle T^{*}x,T^{*}x \rangle = \langle x,TT^{*}x \rangle = \langle x,T^{*}Tx \rangle = \langle x,x \rangle$

This certainly holds for full isometries, since they are normal. Is this true for partial isometries, when applied to the domain on which they preserve norms? Specifically, if $x \in W$ then is it true that $TT^{*}x = T^{*}Tx$? Is there a way to arrive at this from first principles?

Answer 1

Notice that your subspace $W$ is in fact uniquely determined by $T$, namely it must be $W = N(T)^\perp$.

Since $T|_{W^\perp} = 0$ clearly $W^\perp \subseteq N(T)$. Conversely, for $x \in N(T)$ write it as $x = y+z$ with $y \in W$ and $z \in W^\perp$. Then $$0 = \|Tx\| = \|Ty+Tz\| = \|Ty\| = \|y\|$$ so $y = 0$ and hence $x = z \in W^\perp$.

Therefore, the question is if $T$ is isometric on $N(T)^\perp$, must $T^*$ also be isometric on $N(T)^\perp$? To see that this is false in general, consider $$T = \frac1{\sqrt{2}} \begin{bmatrix} 1 & 0 \\ 1 & 0\end{bmatrix}.$$

Answer 2

No, this is not true as the example $T=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix},\ \ x=\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ shows (in matrices form, w.r.t to the usual dot product). $$TT^*=\begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}\ and\ T^*T=\begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} $$ (please check the calculation yourself)