Commuting self adjoint operator exercise.

Question

Let $V$ be a complex vector space with positive definite inner product $(, )$, and $T: V \to V$ a linear map. Recall that the adjoint $T^*$ of $T$ is defined by: $$(T(x), y) = (x, T^*(y))$$ for all $x \in y \in V$, and that $T$ is called normal if $T$ and $T^*$ commute. Suppose that $T^3 = TT^*$. Let $U$ be the kernel of $T$ and $W$ the orthogonal complement of $U$ in $V$.

I've solved the first 3 parts but I'm struggling with the 4th one. I've included them in this post because I think the 4th part uses the first 3.

(1) Show that $W$ is $T^*$-invariant

Let $w \in W$ so that $(w, u) = 0$ for all $u \in U$. Now we have $(T^* w, u) = (w, T^{**}u) = (w, Tu) = (w, 0) = 0$.

(2) Show that $U$ is $T^*$ invariant.

Let $u \in U$ so that $Tu =0$. Then $T(T^*u) = T^3 u = 0$.

(3) Conclude that $W$ is $T$-invariant

Suppose $w \in W$ so that $(w, u) =0$ for all $u \in U$. Then $(Tw, u) = (w, T^*u ) = 0$ where in the last inequality we used part (b).

(4) Show that the restrictions of $T$ and $T^*$ to $W$ commute.

Answer 1

$T(TT^*-T^*T)T^* = 0$ Since $Ker(T) = U \implies Range((TT^*-T^*T)T^*) \subseteq U$ But $T^*$ and $(TT^*-T^*T)$ are $W$-invariant and hence $Ker(TT^*-T^*T) \supseteq T^*(W)$.

If $W$ is finite dimensional then since $T|_W$ is injective and hence surjective, we have that $T^*|_W$ is injective and surjective and hence $T^*(W) = W$. Since $Ker(TT^*-T^*T) \supseteq T^*(W)$, we now have $Ker(TT^*-T^*T) \supseteq W$.

Hence $(TT^*-T^*T)|_W = 0$ and hence $TT^*=T^*T$ when restricted to $W$.

Answer 2

Following Brian's suggestion, let's examine the restrictions to $W$, $T|_W \colon W \to W$ and $T^*|_W \colon W \to W$. It is parts (1) and (3) that show that these restrictions make sense.

Let $w \in W$. Then by the invariance of $T|_W$ and $T^*|_W$, we have that $(TT^* - T^* T)w \in W$. Observe, $$\begin{align} T (TT^* - T^* T) &= T(T T^*) - (T T^*) T\\ &= TT^3 - T^3 T\\ &= 0\\ \Rightarrow T (TT^* - T^* T) w &= 0\\ \Rightarrow (TT^* - T^* T) w &\in U \quad (= \ker T) \end{align}$$ It follows that $(TT^* - T^* T)w \in W \cap U = \{0\}$, whence $(TT^* - T^* T) w = 0$, as desired.