$T: V \rightarrow V$ is a linear operator, and $TT^*=T^3$, $\text{Ker}(T) = U, W=U^⊥$, Prove that $T_{|_W}$ is normal

Question

I proved that $W$ is $T^*-\text{invariant}$, and that $W$ is $T-\text{invariant}$.

I couldn't find a good way to prove that the operator $T_{|_W}$ is normal.

here's what I did so far:

$U=\text{Ker}(T) \Rightarrow U \cap W = \{0\} \Rightarrow \text{Ker}(T_{|_W}) = \{0\} \Rightarrow T_{|_W} \text{is one-to-one}$

I tried to prove it because there was a hint saying: "first show that $T_W$ is one-to-one".

Thanks guys :)

Edit: forgot to mention that $U \cap W = \{0\}$ comes from the fact that $V=U\oplus W$

Answer 1

If $W$ is a subspace of $V$ which is both $T$-invariant and $T^{*}$ invariant then $\left( T|_{W} \right)^{*} = T^{*}|_{W}$. This follows from the defining property of the adjoint operator. Now, in your case, $T|_{W} \colon W \rightarrow W$ is one-to-one, hence onto, hence invertible. Given $w \in W$, we have

$$ \left( T|_{W} \circ T^{*}|_{W} \right)(w) = (T \circ T^{*})(w) = T^3(w) = \left( T|_{W} \right)^3(w) \implies \\ T^{*}|_{W}(w) = \left( \left( T|_{W} \right)^{-1} \circ \left( T|_{W} \right)^3 \right)(w) = \left( T|_{W} \right)^2 (w).$$

Hence,

$$ \left( T|_{W} \right)^{*} \circ T|_{W} = T^{*}|_{W} \circ T|_{W} = \left( T|_{W} \right)^2 \circ T|_{W} = \left( T|_{W} \right)^3, \\ T|_{W} \circ \left( T|_{W} \right)^{*} = T|_{W} \circ T^{*}|_{W} = T|_{W} \circ \left( T|_{W} \right)^2 = \left( T|_{W} \right)^3 $$

and so $T|_{W}$ is normal.

Answer 2

$TT^*=T^3$
$\implies T(TT^*) = (TT^*)T\implies (T^*T)(TT^*)= T^*(TT^*)T=(T^*T)^2$
finish 1
since $W$ is $T$ and $T^*$ invariant and $T$ is injective when restricted to $W$, being finite dimensional, $T$ is invertible when restricted to $W$ which implies $(T_{\vert W}^*T_{\vert W})$ is invertible which implies $(T_{\vert W}^*T_{\vert W}) = (T_{\vert W}T_{\vert W}^*)$
so $T$ is normal when restricted to $W$

finish 2
working directly on $V$ and using $(T^*T)(TT^*)=(T^*T)^2$
taking trace of each side we have
$\text{trace}\Big(\big(T^*T\big)^2\Big) =\Big \Vert T^*T\Big \Vert_F\Big \Vert TT^*\Big \Vert_F\geq \text{trace}\Big(\big(T^*T\big)\big(TT^*\big)\Big) = \text{trace}\Big(\big(T^*T\big)^2\Big)$
thus Cauchy-Schwarz is met with equality and
$\big(T^*T\big)= \eta \cdot \big(TT^*\big)$
taking the trace of each side once more and (ignoring the case of $T=\mathbf 0$ which trivially holds), we see $\eta =1\implies \big(T^*T\big)= \big(TT^*\big)$
so $T$ is normal. And this necessarily holds when restricting to any $T$-invariant subspace.