Prove null $T^k$ = null $T$ and range $T^k$ = range $T$

I'm trying to prove that if $T$ is a normal operator, then null $T^k$ = null $T$ and range $T^k$ = range $T$. Showing null $T$ $\subset$ null $T^k$ is simple, so I'm working on the other inclusion. So far I've been able to deduce that for a vector $v \in$ null $T^k$ we have $TT^\star v = T^\star Tv$ $\implies$ $T^k T^\star v = T^\star T^k v$ $\implies$ $T^k T^\star v = 0$ $\implies$ $T^\star v \in$ null $T^k$. I'm not sure if this is useful though, and I'm stuck on where I should go from here.

• Do you have an inner-product floating around to use?
– Tom
Nov 20, 2013 at 23:47
• I tried constructing some using various combinations of operators such as $T^{k-1} T^\star T$, but I couldn't find one that was very helpful. Nov 20, 2013 at 23:51
• By $T^k$ do you mean the $T$ to the $k$-th power? Nov 20, 2013 at 23:51
• @Hayden yes I do Nov 20, 2013 at 23:52
• @Danny Are you working over finite dimensions?
– Tom
Nov 21, 2013 at 0:07

First for $$S$$ self-adjoint: suppose $$S^kx = 0$$. Then

$$0 = \langle S^{k}x, S^{k-2}x \rangle = \langle S^{k-1}x, S^{k-1}x\rangle$$

so by positive definiteness of inner product, $$S^{k-1}x = 0$$, and we can continue down to $$Sx=0$$.

If $$T$$ is normal, suppose $$T^kx=0$$. Then

$$(T^*T)^kx = (T^*)^k (T^kx) = 0$$

(The key is $$(T^*T)^k = (T^*)^k T^k$$ since $$T$$ is normal). So by the first part (since $$T^*T$$ is self adjoint)

$$0 = \langle T^* Tx, x \rangle = \langle Tx, Tx \rangle$$

so $$Tx = 0$$.

To show $$\operatorname{Rg}(T^k)=\operatorname{Rg}(T)$$, note using the above result,

$$\operatorname{Rg}(T^k) = \operatorname{Ker}((T^k)^*)^\perp = \operatorname{Ker}((T^*)^k)^\perp\\ = \operatorname{Ker}(T^*)^\perp = \operatorname{Rg}(T).$$

Comment: this says that any normal operator has the same kernel as any of its powers. If $$T$$ is normal, then $$T-\lambda I$$ is normal, which shows that $$(T-\lambda I)^kx = 0 \Rightarrow (T-\lambda I)x=0$$. This shows that any normal operator in finite dimensions is diagonalizable over $$\mathbb{C}$$. Some short additional work is needed to show it is orthogonally diagonalizable.

• Nicely done. Nitpick: if $T^kx=0$, then $(T^*T)^kx=(T^*)^kT^kx=0$. Hence $T^*Tx=0$ by the first part etc... You only said that $\langle (T^*T)^kx,x\rangle =0$, which does not obviously imply $(T^*T)^kx=0$. Nov 21, 2013 at 2:41
• @Danny First question: the definition of the adjoint is that $\langle Ax, x \rangle = \langle x, A^*x \rangle$. But by assumption, $S^* = S$. So just move one of the $S$ over to the right side. Second question: two well-known properties of the adjoint are that $(AB)^* = B^* A^*$ and $(A^{*})^* = A$. Therefore $(T^* T)^* = T^*(T^{*})^{*} = T^*T$. You can find this material in any linear algebra book. For a first text I recommend Strang's Linear Algebra and its Applications; for a more advanced text I recommend Roman's Advanced Linear Algebra. Nov 24, 2013 at 19:13
• @Danny More explicitly, $\langle S^kx, S^{k-2}x\rangle = \langle S(S^{k-1}x), S^{k-2}x \rangle = \langle S^{k-1}x, S^*(S^{k-2}x)\rangle = \langle S^{k-1}x, S(S^{k-2}x) \rangle = \langle S^{k-1}x, S^{k-1}x \rangle$. Nov 24, 2013 at 19:19
• Thanks for the answer. I suppose T^k=0 should be T^kx = 0, I could not edit due to minimum character limit. Nov 23, 2018 at 20:36
• @EricAuld If T is normal, suppose Tk=0. Then ... May 14, 2019 at 11:39

Let $T^{k}v=0$. Then

$(T^*T)^kv=0$ and

$[(T^*T)^\dagger T^*T]^kx=0$.

Then $T^*Tx=0$ and so $Tx=0$, premultiplying both sides with Moore-Penrose inverse of $T^*$.

(Note that Moore-Penrose inverse of a self adjoint element commutes with that element.)

I'm not sure this is correct.

$$T$$ is normal $$\implies$$ $$\ker T = \ker T^*$$ $$\implies$$ $$\ker T = (\operatorname{range} T) ^\perp$$.

Now use induction: $$\ker T^{k+1} = \{v: Tv \in \ker T^k\} = \{v: Tv \in \ker T\}$$ (by the induction hypothesis).

So, $$v \in \ker T^{k+1}$$ implies that $$Tv \in \ker T$$; since $$Tv \in \operatorname{range} T$$ trivially, and $$\ker T \perp \operatorname{range} T$$, it follows that $$Tv=0$$, i.e. $$v \in \ker T$$. So $$\ker T^{k+1} \subseteq \ker T$$, the opposite inclusion being trivial.