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In the context of a Hilbert space $H$, when an operator $A$ is diagonalizable we usually decompose the Hilbert space into direct sum of eigenspaces $$H=\bigoplus\limits_{n=1}^\infty E_n$$ where $E_n$ denotes the n$th$ eigenspace of $A$.

I am wondering why we don't write $$H=\overline{\bigoplus\limits_{n=1}^\infty E_n}$$ Are we sure that all the eigenspaces $E_n$ are closed subspaces of $H$? Why is it so?

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An eigenspace of a bounded operator $A$ is closed, because it is the kernel of $A-\lambda I$. More generally, an eigenspace of a closed operator $A$ is closed (the proof is not hard, but seems tangential to this question – it begins by considering the intersection of the graph of $A$ and the graph of $\lambda I$, which is clearly closed, and then projecting that onto its first component).

As to why we don't put a bar above the direct sum, well you would have to do that if you defined the direct sum as the algebraic direct sum (which is just the finite sums of vectors from different $E_n$). But one rarely bothers with the algebraic direct sum, so the bar would just get in the way. Especially when the spaces are orthogonal, so each member of the direct sum can really be written uniquely as a convergent sum $\sum_{n=1}^\infty v_n$ with $v_n\in E_n$ for each $n$.

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  • $\begingroup$ Thanks. "An eigenspace of a closed operator A is closed" is really what I was looking for. Could you provide a few more details as for why that is true? $\endgroup$
    – Yun
    Dec 5, 2013 at 17:03
  • $\begingroup$ The point is this: The projection of the graph of $\lambda I$ on the first coordinate (i.e., $(x‚\lambda x)\mapsto x$) is an isomorphism. It is clearly bounded, and the inverse $x\mapsto(x,\lambda x)$ is bounded as well. Thus a closed subspace is mapped to a closed subspace, and that's all. $\endgroup$ Dec 5, 2013 at 23:21

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