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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.

Danny
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3 Answers3

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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.

Eric Auld
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  • 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 $(T^*T)^kx=0$. – Julien Nov 21 '13 at 02:41
  • Quite right, I appreciate the nitpick actually. That is an important detail. Edited. – Eric Auld Nov 21 '13 at 03:45
  • Sorry for the late response. Could you explain why $\langle S^k x, S^{k-2} x \rangle = \langle S^{k-1} x, S^{k-1} x \rangle$? Also, how do you know that $T^\star T$ is self-adjoint? – Danny Nov 24 '13 at 18:38
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    @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. – Eric Auld Nov 24 '13 at 19:13
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    @Danny More explicitly, $\langle S^kx, S^{k-2}x\rangle = \langle S(S^{k-1}x), S^{k-2}x \rangle = \langle S^{k-1}x, S^*(S^{k-2}x)\rangle = \langle S^{k-1}x, S(S^{k-2}x) \rangle = \langle S^{k-1}x, S^{k-1}x \rangle$. – Eric Auld Nov 24 '13 at 19:19
  • I understand it now. Thanks! – Danny Nov 24 '13 at 20:19
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    Thanks for the answer. I suppose T^k=0 should be T^kx = 0, I could not edit due to minimum character limit. – unlut Nov 23 '18 at 20:36
  • @unlut Where do you mean? – Eric Auld May 14 '19 at 11:36
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    @EricAuld If T is normal, suppose Tk=0. Then ... – unlut May 14 '19 at 11:39
  • You're right. Thanks – Eric Auld May 14 '19 at 15:36
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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.

Alex M.
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Michele Monaco
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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.)

AnyAD
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