Finite-dimensional subalgebra of $\mathbf{B}(H)$ is a von Neumann algebra

Question

Let $H$ be a complex Hilbert-space and $\mathbf{B}(H)$ the space of bounded linear operators $H \to H$.

The first example of a von Neumann algebra in the book I'm reading is: any finite-dimensional unital $*$-subalgebra $A$ of $\mathbf{B}(H)$, which is finite-dimensional.

I've tried to prove this. My idea was maybe bit of an overkill: Let $P_A : \mathbf{B}(H) \to A$ be the projection onto $A$. By Kaplansky's Theorem, the image of a ultraweakly continuous unital $*$-homomorphism $M \to \mathbf{B}(K)$ (for $M$ a von Neumann algebra and $K$ a Hilbert-space) is a von Neumann algebra.

So I need to show that for any ultraweakly continuous functional $\varphi : A \to \mathbf{C}$ the composition $\varphi \circ P_A$ is ultraweakly continuous. This would follow if $\Vert P_A \Vert \leq 1$, which I wasn't able to show.

Indeed, if $x_i \in (\mathbf{B}(H))_1$ is a converging net $x_i \to x$ in the strong operator topology, then $\Vert (P_A x_i - P_Ax) \xi \Vert \leq \Vert (x_i - x) \xi \Vert \to 0$ for all $\xi \in H$. Thus $\varphi \circ P_A (x_i) \to \varphi\circ P_A (x)$ as $\varphi$ is strongly continuous on $(V)_1$.

I have two questions: Is it true that $\Vert P_A \Vert \leq 1$ for any finite-dimensional $*$-subalgebra? And I feel like there should be a more direct way to prove the fact than using Kaplansky. How can I do it?

Thanks!

Answer 1

For any finite-dimensional space all its Hausdorff vector topologies are equivalent. This is proved in Functional analysis by W. Rudin.

In particular, any finite-dimensional subspace (algebra structure is irrelevant here), say $A$, of $ \mathcal{B}(H)$ is closed in any standard operator topology, e.g. weak topology.

Answer 2

You could also appeal to the classification of finite-dimensional C$^\ast$-algebras as well: They are all of the form $$\mathbb{M}_{n_1}(\mathbb{C}) \oplus \mathbb{M}_{n_2}(\mathbb{C}) \oplus \mathbb{M}_{n_3}(\mathbb{C}) \oplus \ldots \oplus \mathbb{M}_{n_k}(\mathbb{C})$$ with $k$ being some finite positive integer. Being a finite direct summand of von Neumann algebras, it is a von Neumann algebra itself (strong/weak-operator convergence occurs componentwise).