# What does "operators on a Hilbert space form an algebra" mean?

I was reading some group theory notes and I am familiar with the concept of a Lie algebra, but I cannot imagine what the following formulation means:

What is more, not only states, but also the operators come in representations of the symmetry group. Clearly, the operators on a Hilbert space form an algebra, hence in particular a vector space.

Does this refer to the eigenspace of the operator?

• It's an algebra, ie it forms a vector space (you can add and scale operators), and there exists a distributive product on them (performing successive applications of operators on the Hilbert vector). Sep 28, 2021 at 10:05
• Wikipedia Sep 28, 2021 at 10:15
• Ah ok thanks thant makes sense! Sep 28, 2021 at 10:18

As per the comments, an algebra $$(\mathcal A, \circ)$$ consists of a vector space $$\mathcal A$$ and a bilinear operation $$\circ:\mathcal A\times \mathcal A \rightarrow \mathcal A$$ which provides some notion of vector multiplication. Examples include $$\mathbb R^3$$ equipped with the cross product and the space of $$2\times 2$$ skew-hermitian matrices equipped with the commutator bracket.
If $$\mathcal H$$ is some Hilbert space, then the bounded linear operators $$\mathfrak B(\mathcal H)$$ also constitute an algebra, with the multiplication operation being given by composition. In other words, for $$\alpha,\beta\in\mathfrak B(\mathcal H)$$ and $$\psi\in \mathcal H$$,
$$(\alpha \circ \beta)(\psi):= \alpha \big( \beta(\psi)\big)$$