Math 202B: Lecture 6

Let $\mathcal{L}(V)$ be the linear algebra of a Hilbert space $V$ of dimension at least two. We have seen that there does not exist an algebra homomorphism $\mathcal{L}(V) \to \mathbb{C}$ . This suggests that the commutativity index of $\mathcal{L}(V)$ is very low, or equivalently that its noncommutativity index is very high. To make this precise, we want to determine the center of $\mathcal{L}(V).$

Definition 6.1. Given a subalgebra $\mathcal{A}$ of $\mathcal{L}(V)$ , a subspace $W$ of $V$ is said to be $\mathcal{A}$ –invariant if for every $A \in \mathcal{A}$ and every $w \in W$ we have $Aw \in W.$

Proposition 6.1. If $W \leq V$ is an $\mathcal{L}(V)$ -invariant subspace, then either $W=\{0_V\}$ or $W=V.$

Problem 6.1. Prove Proposition 6.1.

Theorem 6.2. The center of $\mathcal{L}(V)$ is the set $\mathbb{C}I$ of scalar operators.

Proof: Let $Z \in \mathcal{L}(V)$ be a central element, and let $\zeta \in \mathrm{Spec}(Z)$ be an eigenvalue of $Z.$ By definition, this means that $\zeta I-Z$ is not invertible in $\mathcal{L}(V),$ which is equivalent to saying that $\mathrm{Ker}(\zeta I-Z)$ is a nonzero subspace of $V$ . We claim that, because $Z$ is a central element in $\mathcal{L}(V),$ the $\zeta$ -eigenspace of $Z$ is $\mathcal{L}(V)$ -invariant. Indeed, for any $v \in \mathrm{Ker}(\zeta I-Z)$ we have

$ZAv = AZv =\zeta Av,$

which shows that $Av$ is again in the $\zeta$ -eigenspace of $Z.$ Since this space has positive dimension, Proposition 6.1 forces $\mathrm{Ker}(\zeta I-Z)=V,$ so that $Zv = \zeta v$ for every $v \in V.$

-QED

Having understood the center of $\mathcal{L}(V),$ let us consider arbitrary commutative subalgebras of $\mathcal{L}(V).$ These can be understood using the Spectral Theorem from Math 202A, which may be stated as follows.

Theorem 6.3. If $A_1,\dots,A_m \in \mathcal{L}(V)$ are commuting normal operators, then there exists an orthonormal basis $X \subset V$ such that $A_1,\dots,A_m \in \mathcal{D}(X).$

Since $\mathcal{D}(X)$ is isomorphic to $\mathcal{F}(X)$ , we can combine the above formulation of the Spectral Theorem with the characterization of commutative algebras obtained in Lecture 1 and the classification of subalgebras of function algebras obtained in Lecture 3 to prove the following.

Theorem 6.4. Every commutative subalgebra of $\mathcal{L}(V)$ is isomorphic to a function algebra.

Proof: Let $\mathcal{A}$ be a subalgebra of $\mathcal{L}(V).$ Since $\mathcal{A}$ is a vector space, it has a basis $A_1,\dots,A_m,$ where $1 \leq m \leq (\dim V)^2.$ If $\mathcal{A}$ is commutative, all of its elements are normal (Theorem 1.3 in Lecture 1), and the basis $A_1,\dots,A_m$ of $\mathcal{A}$ consists of commuting normal operators. By Theorem 6.3, there exists an orthonormal basis $X \subset V$ such that $A_1,\dots,A_m \in \mathcal{D}(X)$ , and therefore $\mathcal{A}$ is a subalgebra of $\mathcal{D}(X)$ . Since $\mathcal{D}(X)$ is isomorphic to the function algebra $\mathcal{F}(X),$ and since every subalgebra of $\mathcal{F}(X)$ is isomorphic to a function algebra (Lecture 3), it follows that $\mathcal{A}$ is isomorphic to a function algebra.

In fact, we can be more precise (though we don’t need to be): $\mathcal{A}$ is isomorphic to $\mathcal{F}(\mathbf{P})$ for some partition $\mathbf{P}$ of $X,$ and we can describe this partition in terms of the eigenspaces of the commuting normal operators $A_1,\dots,A_m$ . Namely, each of the operators $A_i$ , $1 \leq i \leq m$ induces a partition $\mathbf{P}_i$ of $X$ such that $x,y \in X$ belong to the same block of $\mathbf{P}_i$ if and only if they belong to the same eigenspace of $A_i.$ Letting $\mathbf{P}=\max(\mathbf{P}_1,\dots,\mathbf{P}_m)$ be the coarsest partition of $X$ finer than all of the $\mathbf{P}_i$ ‘s, a little thought reveals that $\mathcal{A}$ is isomorphic to $\mathcal{F}(\mathbf{P}).$

-QED

As we have just seen, for any commutative subalgebra $\mathcal{A}$ of $\mathcal{L}(V)$ the spectral theorem guarantees a corresponding decomposition of $V$ into a direct sum of orthogonal one-dimensional subspaces on which $\mathcal{A}$ acts diagonally. If $\mathcal{A}$ is a noncommutative subalgebra of $\mathcal{L}(V)$ , this is no longer the case. Nevertheless, we can consider the set of all $\mathcal{A}$ -invariant subspaces of $V,$ which is an induced sublattice of the lattice of all subspaces of $V$ , since the intersection of two $\mathcal{A}$ -invariant subspaces is again $\mathcal{A}$ -invariant, and likewise the span of the union of two $\mathcal{A}$ -invariant subspaces is again $\mathcal{A}$ -invariant (exercise: check these statements). For any subalgebra $\mathcal{A} \leq \mathcal{L}(V)$ , the lattice of $\mathcal{A}$ -invariant subspaces contains at least $\{0_V\}$ and $V.$ An important theorem of Burnside characterizes subalgebras of $\mathcal{L}(V)$ for which this minimum is achieved.

Theorem 6.5. (Burnside) A subalgebra $\mathcal{A}$ of $\mathcal{L}(V)$ has exactly two invariant subspaces if and only if $\mathcal{A}=\mathcal{L}(V).$

Proof: An elementary proof of Burnside’s theorem by induction in the dimension of $V$ can be found here.

-QED

Corollary 6.6. A proper subalgebra $\mathcal{A}$ of $\mathcal{L}(V)$ has at least four invariant subspaces.

Proof: The zero space and the total space are $\mathcal{A}$ -invariant, and since $\mathcal{A} \neq \mathcal{L}(V)$ there is a third $\mathcal{A}$ -invariant subspace $\{0_V\} < W < \mathcal{L}(V),$ by Burnside’s theorem. The orthogonal complement $W^\perp$ of $W$ is again a nonzero proper subspace of $V,$ and we claim it is $\mathcal{A}$ -invariant. Indeed, let $x \in W^\perp$ and $A \in \mathcal{A}.$ Then, for any $w \in W$ we have

$\langle w,Ax \rangle = \langle A^*w,x \rangle = 0,$

where we used the fact that $W$ is $\mathcal{A}$ -invariant. This shows $Ax \in W^\perp,$ i.e. $W^\perp$ is $\mathcal{A}$ -invariant.

– QED

We can attempt to classify subalgebras of $\mathcal{L}(V)$ by adapting our approach to the classification of subalgebras of $\mathcal{F}(X),$ where we looked at subalgebras of $\mathcal{F}(X)$ corresponding to partitions of the set $X.$ To emulate this approach, we need the Hilbert space analogue of a set partition, in which “nonempty” is replaced with “nonzero” and “disjoint” becomes “orthogonal.”

Definition 6.2. A linear partition of $V$ is a set $\mathbf{W}$ of nonzero pairwise orthogonal subspaces of $V$ whose union spans $V.$ We refer to the subspaces $W \in \mathsf{W}$ as the blocks of $\mathbf{W},$ and write

$V = \bigoplus\limits_{W \in \mathsf{W}} W.$

Given a linear partition $\mathbf{W}$ of $V,$ we define $\mathcal{A}(\mathbf{W})$ to be the set of all operators $A \in \mathcal{L}(V)$ such that every block $W \in \mathbf{W}$ is $A$ -invariant.

Problem 6.2. Prove that $\mathcal{A}(\mathbf{W})$ is a subalgebra of $\mathcal{L}(V).$

Here our construction starts to diverge from what we saw in the setting of function algebras: unless $\mathbf{W}=\{V\}$ is the linear partition of $V$ consisting of a single block, in which case $\mathcal{A}(\mathbf{W})=\mathcal{L}(V),$ the subalgebra $\mathcal{A}(\mathbf{W})$ is definitely not isomorphic to the algebra of all linear operators on a Hilbert space. Indeed, consider the case where $\mathbf{W}=\{W_1,W_2\}$ is a linear partition of $V$ with two blocks, i.e. $V = W_1 \oplus W_2$ . Let $e_1,\dots,e_m$ be an ordered basis of $W_1,$ and let $f_1,\dots,f_n$ be an ordered basis of $W_2.$ Then $e_1,\dots,e_m,f_1,\dots,f_n$ is an ordered basis of $V$ , and $\mathcal{A}(\mathbf{W})$ consists of all operators in $A\in \mathcal{L}(V)$ whose matrix relative to this ordered basis has the form

$[A]_{(e_1,\dots,e_m,f_1,\dots,f_n)}=\begin{bmatrix} M_1 & {} \\ {} & M_2 \end{bmatrix},$

with $M_1 \in \mathbb{C}^{m \times m}$ and $M_2 \in \mathbb{C}^{n\times n}.$ Thus, $\mathcal{A}(\{W_1,W_2\})$ is not isomorphic to a linear algebra, but rather to a direct sum of linear algebras,

$\mathcal{A}(\{W_1,W_2\}) = \mathcal{L}(W_1) \oplus \mathcal{L}(W_2).$

The question now is whether the above can reversed: given a subalgebra $\mathcal{A}$ of $\mathcal{L}(V),$ does there exist a linear partition $\mathbf{W}$ such that $\mathcal{A}=\mathcal{A}(\mathbf{W})$ ? If $\mathcal{A}=\mathcal{L}(V),$ the answer is clearly “yes” since we have $\mathcal{L}(V) = \mathcal{A}(\{V\}).$ Thus we consider the case where $\mathcal{A}$ is a proper subalgebra of $\mathcal{L}(V).$ Our starting point is Burnside’s theorem, which guarantees the existence of an $\mathcal{A}$ -invariant subspace $\{0_V\} < W < V$ . Together with the corollary to Burnside’s theorem, we thus have a linear partition $\mathbf{W} = \{W,W^\perp\}$ of $V$ whose blocks are $\mathcal{A}$ -invariant. We now ask whether $W$ has a proper non-trivial subspace invariant under $\mathcal{A}$ . If it does we can split it into the orthogonal direct sum of two smaller $\mathcal{A}$ -invariant subspaces; if not $W$ is said to be $\mathcal{A}$ –irreducible. Continuing this process, for both $W$ and $W^\perp$ , we get the following.

Theorem 6.7. (Maschke’s theorem) There exists a linear partition $\mathbf{W}$ of $V$ whose blocks are $\mathcal{A}$ -invariant and $\mathcal{A}$ -irreducible subspaces.

Now consider the linear partition $\mathbf{W}$ of $V$ into $\mathcal{A}$ -invariant, $\mathcal{A}$ -irreducible subspaces we have constructed. It is tempting to hope that $\mathcal{A} = \mathcal{A}(\mathsf{W}).$ Unfortunately, this is not quite correct. For example, consider the case where $V$ is a Hilbert space of even dimension $2n.$ Let $X=\{y_1,\dots,y_n,z_1,\dots,z_n\}$ be an othornormal basis of $V$ , and consider the set $\mathcal{A}$ of all operators $A \in \mathcal{L}(V)$ such that

$\langle x_i,Ax_j \rangle = \langle y_i,Ay_j \rangle, \quad 1 \leq i,j \leq N.$

Then $\mathcal{A}$ is a subalgebra of $\mathcal{L}(V)$ – it consists of all operators whose matrix in the basis $X$ is block diagonal with two identical blocks. Now $V=W \oplus W^\perp$ , where $W= \mathrm{Span}\{x_1,\dots,x_n\}$ and $W^\perp = \mathrm{Span}\{y_1,\dots,y_n\},$ and this is a linear partition of $V$ into $\mathcal{A}$ -invariant and $\mathcal{A}$ -irreducible subspaces. However, the algebra $\mathcal{A}(W \oplus W^\perp)$ is not $\mathcal{A}$ — it is the larger algebra of operators whose matrix relative to $X$ is block diagonal, but not necessarily with both blocks being the same. In other words $\mathcal{A}$ consists of all matrices of the form

$\begin{bmatrix} A & {} \\ {} & A \end{bmatrix},$

whereas $\mathcal{A}(W \oplus W^\perp)$ consists of all matrices of the form

$\begin{bmatrix} A_1 & {} \\ {} & A_2 \end{bmatrix}.$

To account for the above situation, we need a notion of equivalence for $\mathcal{A}$ -invariant subspaces of $V$ .

Definition 6.3. We say that two $\mathcal{A}$ -invariant subspaces $W_1,W_2$ of $V$ are $\mathcal{A}$ -equivalent if we can choose bases in $W_1$ and $W_2$ such that, for every $A \in \mathcal{A},$ the matrix of the restriction of $A$ to $W_1$ is the same as the matrix of the restriction of of $A$ to $W_2.$

Now let $\Lambda$ be a set parameterizing equivalence classes of $\mathcal{A}$ -invariant and $\mathcal{A}$ -irreducible subspaces of $V.$ For each $\lambda \in \Lambda,$ let $V^\lambda$ be a representative of the the equivalence class corresponding to $\lambda.$ The following result is a very simple version of the Artin-Wedderburn theorem. We will eventually prove this theorem for a special class of algebras coming from finite groups.