invariant subspaces for self-adjoint *-algebras of operators

In this entry we provide few results concerning invariant subspaces of *-algebras of bounded operators on Hilbert spaces.

Let $H$ be a Hilbert space and $B(H)$ its algebra of bounded operators. Recall that, given an operator $T\in B(H)$, a subspace $V\subseteq H$ is said to be invariant for $T$ if $Tx\in V$ whenever $x\in V$.

Similarly, given a subalgebra $𝒜\subseteq B(H)$, we will say that a subspace $V\subseteq H$ is invariant for $𝒜$ if $Tx\in V$ whenever $T\in 𝒜$ and $x\in V$, i.e. if $V$ is invariant for all operators in $𝒜$.

Invariant subspaces for a single operator

Proposition 1 - Let $T\mathrm{\in }B\mathit{}\mathrm{(}H\mathrm{)}$. If a subspace $V\mathrm{\subset }H$ is invariant for $T$, then so is its closure $\overline{V}$.

Proof: Let $x\in \overline{V}$. There is a sequence $\{x_n\}$ in $V$ such that $x_n\to x$. Hence, $T{x}_n\to Tx$. Since $V$ is invariant for $T$, all $T{x}_n$ belong to $V$. Thus, their limit $Tx$ must be in $\overline{V}$. We conclude that $\overline{V}$ is also invariant for $T$. $\mathrm{\square }$

$$

Proposition 2 - Let $T\mathrm{\in }B\mathit{}\mathrm{(}H\mathrm{)}$. If a subspace $V\mathrm{\subset }H$ is invariant for $T$, then its orthogonal complement $V^{\mathrm{⟂}}$ is invariant for $T^{\mathrm{*}}$.

Proof: Let $y\in V^{⟂}$. For all $x\in H$ we have that $⟨x,T^{*}y⟩=⟨Tx,y⟩=0$, where the last equality comes from the fact that $Tx\in V$, since $V$ is invariant for $T$. Therefore $T^{*}y$ must belong to $V^{⟂}$, from which we conclude that $V^{⟂}$ is invariant for $T^{*}$. $\mathrm{\square }$

$$

Proposition 3 - Let $T\mathrm{\in }B\mathit{}\mathrm{(}H\mathrm{)}$, $V\mathrm{\subset }H$ a closed subspace and $P\mathrm{\in }B\mathit{}\mathrm{(}H\mathrm{)}$ the orthogonal projection onto $V$. The following are statements are equivalent:

1. $V$ is invariant for $T$.

2. $V^{\mathrm{⟂}}$ is invariant for $T^{\mathrm{*}}$.

3. $T\mathit{}P\mathrm{=}P\mathit{}T\mathit{}P$.

Proof: $(1)⟹(2)$ This part follows directly from Proposition 2.

$(2)⟹(1)$ From Proposition 2 it follows that ${(V^{⟂})}^{⟂}$ is invariant for ${(T^{*})}^{*}=T$. Since $V$ is closed, $V=\overline{V}={(V^{⟂})}^{⟂}$. We conclude that $V$ is invariant for $T$.

$(1)⟹(3)$ Let $x\in H$. From the orthogonal decomposition theorem we know that $H=V\oplus V^{⟂}$, hence $x=y+z$, where $y\in V$ and $z\in V^{⟂}$. We now see that $TPx=Ty$ and $PTPx=PTy=Ty$, where the last equality comes from the fact that $Ty\in V$. Hence, $TP=PTP$.

$(3)⟹(1)$ Let $x\in V$. We have that $Tx=TPx=PTPx$. Since $PTPx$ is obviously on the image of $P$, it follows that $Tx\in V$, i.e. $V$ is invariant for $T$. $\mathrm{\square }$

$$

Proposition 4 - Let $T\mathrm{\in }B\mathit{}\mathrm{(}H\mathrm{)}$, $V\mathrm{\subset }H$ a closed subspace and $P\mathrm{\in }B\mathit{}\mathrm{(}H\mathrm{)}$ the orhtogonal projection onto $V$. The subspaces $V$ and $V^{\mathrm{⟂}}$ are both invariant for $T$ if and only if $T\mathit{}P\mathrm{=}P\mathit{}T$.

Proof: $(⟹)$ From Proposition 3 it follows that $V$ is invariant for both $T$ and $T^{*}$. Then, again from Proposition 3, we see that $PT={(T^{*}P)}^{*}={(P{T}^{*}P)}^{*}=PTP=TP$.

$(⟸)$ Suppose $TP=PT$. Then $PTP=TPP=TP$, and from Proposition 3 we see that $V$ is invariant for $T$.

We also have that $P{T}^{*}=T^{*}P$, and we can conclude in the same way that $V$ is invariant for $T^{*}$. From Proposition 3 it follows that $V^{⟂}$ is also invariant for $T$. $\mathrm{\square }$

Invariant subspaces for *-algebras of operators

We shall now generalize some of the above results to the case of self-adjoint subalgebras of $B(H)$.

Proposition 5 - Let $\mathrm{A}$ be a *-subalgebra of $B\mathit{}\mathrm{(}H\mathrm{)}$ and $V$ a subspace of $H$. If a subspace $V$ is invariant for $\mathrm{A}$, then so are its closure $\overline{V}$ and its orthogonal complement $V^{\mathrm{⟂}}$.

Proof: From Proposition 1 it follows that $\overline{V}$ is invariant for all operators in $𝒜$, which means that $V$ is invariant for $𝒜$.

Also, from Proposition 2 it follows that $V^{⟂}$ is invariant for the adjoint of each operator in $𝒜$. Since $𝒜$ is self-adjoint, it follows that $V^{⟂}$ is invariant for $𝒜$. $\mathrm{\square }$

$$

Theorem - Let $\mathrm{A}$ be a *-subalgebra of $B\mathit{}\mathrm{(}H\mathrm{)}$, $V\mathrm{\subset }H$ a closed subspace and $P$ the orthogonal projection onto $V$. The following are equivalent:

1. $V$ is invariant for $\mathrm{A}$.

2. $V^{\mathrm{⟂}}$ is invariant for $\mathrm{A}$.

3. $P\mathrm{\in }{\mathrm{A}}^{\mathrm{\prime }}$, i.e. $P$ belongs to the commutant of $\mathrm{A}$.

Proof: $(1)⟺(2)$ This equivalence follows directly from Proposition 5 and the fact that $V$ is closed.

$(1)⟹(3)$ Suppose $V$ is invariant for $𝒜$. We have already proved that $V^{⟂}$ is also invariant for $𝒜$. Thus, from Proposition 4 it follows that $P$ commutes with all operators in $𝒜$, i.e. $P\in {𝒜}^{\prime }$.

$(3)⟹(1)$ Suppose $P\in {𝒜}^{\prime }$. Then $P$ commutes with all operators in $𝒜$. From Proposition 4 it follows that $V$ is invariant for each operator in $𝒜$, i.e. $V$ is invariant for $𝒜$. $\mathrm{\square }$