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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.

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    $\begingroup$ Do you have an inner-product floating around to use? $\endgroup$
    – Tom
    Nov 20, 2013 at 23:47
  • $\begingroup$ 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. $\endgroup$
    – Danny
    Nov 20, 2013 at 23:51
  • $\begingroup$ By $T^k$ do you mean the $T$ to the $k$-th power? $\endgroup$
    – Hayden
    Nov 20, 2013 at 23:51
  • $\begingroup$ @Hayden yes I do $\endgroup$
    – Danny
    Nov 20, 2013 at 23:52
  • $\begingroup$ @Danny Are you working over finite dimensions? $\endgroup$
    – Tom
    Nov 21, 2013 at 0:07

3 Answers 3

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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.

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  • $\begingroup$ 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 $(T^*T)^kx=0$. $\endgroup$
    – Julien
    Nov 21, 2013 at 2:41
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    $\begingroup$ @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. $\endgroup$
    – Eric Auld
    Nov 24, 2013 at 19:13
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    $\begingroup$ @Danny More explicitly, $\langle S^kx, S^{k-2}x\rangle = \langle S(S^{k-1}x), S^{k-2}x \rangle = \langle S^{k-1}x, S^*(S^{k-2}x)\rangle = \langle S^{k-1}x, S(S^{k-2}x) \rangle = \langle S^{k-1}x, S^{k-1}x \rangle$. $\endgroup$
    – Eric Auld
    Nov 24, 2013 at 19:19
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    $\begingroup$ Thanks for the answer. I suppose T^k=0 should be T^kx = 0, I could not edit due to minimum character limit. $\endgroup$
    – unlut
    Nov 23, 2018 at 20:36
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    $\begingroup$ @EricAuld If T is normal, suppose Tk=0. Then ... $\endgroup$
    – unlut
    May 14, 2019 at 11:39
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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.)

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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.

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