0
$\begingroup$

I want to prove: given a bounded normal operator $T$ on a Hilbert space $H$. If $H_1$ is a $T$-invariant subspace, then the orthogonal complement $W_1$ of $H_1$ is also $T$-invariant.

I can prove the case when $H$ is finite-dimensional. If $T$ can be written as a matrix $$\begin{pmatrix} M & A\\ 0 & B \end{pmatrix},$$ then by $T^*T=TT^*$,we have $MM^*+AA^*=MM^*$. Take the trace, then we have A must be 0. However, I'm not sure how to translate the language when $H$ is infinite-dimensional where trace might not be well-defined.

$\endgroup$
5
  • 3
    $\begingroup$ Show that $H_1$ is also $T^{\ast}$-invariant. $\endgroup$
    – Daniel Fischer
    Apr 25, 2016 at 15:46
  • $\begingroup$ The only way I can think of proving $H_1$ is $T^*$-invariant is firstly proving $H_1$ has a basis consisting of eigenvectors of $T|_{H_1}$ and then proving any eigenvector of $T$ is also an eigenvector of $T^*$. I know the first argument is true for finite-dimensional $H_1$. But I don't know whether it is so for infinite-dimensional $H_1$. $\endgroup$
    – Fan
    Apr 25, 2016 at 15:53
  • $\begingroup$ @DanielFischer I posted a proof based on results in Rudin's Functional Analysis. Please let me know if there's any simpler proof. $\endgroup$
    – Fan
    Apr 25, 2016 at 19:24
  • $\begingroup$ Impressive proof. Upon thinking about it a bit more, the easiest proof is close to your argument for the finite-dimensional case. Sorry for the misleading comment. We can still write $T$ as a matrix $\begin{pmatrix} M & A \\ 0 & b\end{pmatrix}$, where the entries are operators. With the orthogonal decomposition $H = H_1 \oplus W_1$, we can write every operator $S \colon H \to H$ as a matrix $\begin{pmatrix} U & V \\ W & X\end{pmatrix}$ with $U \colon H_1 \to H_1$, $V \colon W_1 \to H_1$, $W \colon H_1 \to W_1$ and $X \colon W_1 \to W_1$. It is easily checked that then $\endgroup$
    – Daniel Fischer
    Apr 25, 2016 at 20:19
  • $\begingroup$ $S^{\ast} = \begin{pmatrix}U^{\ast} & W^{\ast} \\ V^{\ast} & X^{\ast} \end{pmatrix}$, and the composition of operators corresponds to the multiplication of matrices, where the multiplication of the components is composition of operators. Then, like in the finite-dimensional case computing $T^{\ast} T$ and $TT^{\ast}$, the normality of $T$ leads to the condition $M^{\ast}M = MM^{\ast} + AA^{\ast}$. Since $T(H_1)\subset H_1$, $M$ is normal, and we obtain $AA^{\ast} = 0$. But that implies $A^{\ast} = 0$ and then $A = 0$, which shows that $W_1$ is also $T$-invariant. $\endgroup$
    – Daniel Fischer
    Apr 25, 2016 at 20:19

1 Answer 1

Reset to default
2
$\begingroup$

Let $A$ be the unital commutative $C^*$-algebra generated by $T$, $\Delta$ be the maximal ideal space of $A$, $E$ be the unique resolution of the identity on the Borel subsets of $\Delta$. Write $P_{H_1}:H\rightarrow H_1$ as the projection from $H$ to $H_1$.

For any Borel set $\omega\subset\Delta$, $E(\omega)$ is a projection from $H$ to the range of $E(\omega)$. Thus $P_{H_1}E(\omega)=E(\omega)P_{H_1}$. By Theorem 12.22(e) in Rudin's Functional Analysis, we have $P_{H_1}$ commutes with $T$.

Then $$(T^*H_1,W_1)=(H_1,TW_1)=(P_{H_1}H_1,TW_1)=(P_{H_1}H_1,P_{H_1}TW_1)=(P_{H_1}H_1,TP_{H_1}W_1)=0,$$ thus $T^*H_1\subset H_1$. So $H_1$ is $T^*$-invariant, and the conclusion follows.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .