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Main question:

Let $X$ be a $\mathbb K$-vector space with $\mathbb K=\mathbb R$ or $\mathbb C$. Let $\mathcal L(X)$ be the set of continous linear applications from $X$ to $X$.
Let $\mathcal A$ be a subalgebra of $\mathcal L(X)$ such that $\{b\in\mathcal L(X)\,|\,\forall a\in\mathcal A,\,ab=ba\}=\{\lambda I\,|\,\lambda\in\mathbb K\}\,.$

I am expecting the following result (but do not know how to prove it).

  • Assume $X$ is finite dimensional. Then $\mathcal A=\mathcal L(X)$.

I doubt this is a very original question, could someone indicate me a method to solve this problem?


Some possible extensions:

If it is indeed the case:

  1. Can we generalize this result to the case of $X$ a Hilbert space, i.e. $\overline {\mathcal A}=\mathcal L(X)$ for some topology to be precised on $\mathcal L(X)$?
  2. Same thing for $X$ a Banach space.
  3. And if we replace $\mathcal L(X)$ by a Banach algebra with unity?

1 Answers 1

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The condition you impose on $\mathcal A$ is that its commutant $\mathcal A'$ is trivial.

In the generality you are asking, the result is not true: for example, let $ A=\left\{\begin{bmatrix}a&b\\0&c \end{bmatrix}:\ a,b,c\in\mathbb K \right\}\subset\mathcal L(\mathbb K^2). $ Then $\mathcal A'=\mathbb K\,I$.

Your claim does hold, though, when $\mathcal A$ is a selfadjoint algebra (i.e. contains adjoints of its elements, or transposes if $\mathbb K=\mathbb R$). What you are then looking for is von Neumann Double Commutant Theorem:

If $H$ is a Hilbert space, and $\mathcal A\subset\mathcal L(H)$ is a non-degenerate selfadjoint subalgebra, $ \mathcal A''=\overline{\mathcal A}^{\rm SOT} $ where SOT is the strong operator topology (i.e. pointwise convergence). Other topologies satisfy the equality too.

In the finite-dimensional case, it is easy to see that any selfadjoint subalgebra is closed in the pointwise topology.

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    Thank you very much for your answer.2012-12-05