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I got a small question concerning the proof of a min-max theorem for selfadjoint operators that I'm currently trying to understand.

The article I'm refering to is Section: Compact operators.

So my problem is the proof of showing that the intersection of S' and $S_{k}$ is non-empty. First of all the proof is written down in a bit of confusing way. For example I suspect that S' is basically the space of all eigenvectors belonging to eigenvalues greater or equal to $\lambda_{k}$. (They don't really point out what the $u_{i}$ with i=k,k+1,... are. I suspect that they are the eigenvectors. Nonetheless the notation is a bit misleading since it suggests that there's only one eigenvector $u_{k}$ for the eigenvalue $\lambda_{k}$.) Now the part that puzzles me starts: First, they point out that S' has codimension k-1. Doesn't that mean that we only have one eigenvector for every eigenvalue $\lambda_{i}$ with i

Second, they say that they use a "dimension count argument" to show that the intersection is non-empty. I figured out that argument for the finite dimensional (matrix) case but couldn't figure it out for the infinite dimensional case.

I'm still a student, so I'd be happy if someone could show me just these two arguments in detail. My background is also more from physics, so I'm not an expert when it comes to infinite dimensional vectorspace.

I'd be more than happy if someone could help me. Thanks in advance!!

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Let $W$ be a subspace of a vector space $V$ such that $V/W$ has finite dimension $k-1$, and let $X$ be a subspace of $V$ with dimension $k$. Then $W\cap X\neq \{0\}$.

This follows from the isomorphism $$(X+W)/W\cong X/(X\cap W),$$ because the left-hand side has dimension at most $k-1$, while the right hand side has dimension $k-\dim(X\cap W)$.

This applies to your setting with $V=H$, $W=S'$, and $X=S_k$.

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I'm pretty sure that the $u_k$ are the eigenvectors of the eigenvalues $\lambda_k$. Notice that they list the eigenvalues with their multiplicity, so that you could have, for example, $$ \lambda_1 = 1 $$ $$ \lambda_2 = 1 $$ $$ \lambda_3 = 0.5 $$ etc. if the eigenspace of the eigenvalue with value $1$ has dimension two. It is then understood that you choose two linearly independent eigenvectors that span this eigenspace and call them $u_1$ and $u_2$. So, if the next eigenvalue is $$ \lambda_4 = 0.3 $$ then this means (implied) that the eigenspace of the eigenvalue with value $0.5$ is one dimensional, so that the space $S_4$ has codimension three.

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Hey. I think the part that puzzled me the most was for some reason cut off. So my actual question is how the dimensional argument work in the infinite dimensional case to prove that the insection of S' and $S_{k}$ is non-empty. But thanks for your answer. – Stan Jul 6 '11 at 13:10

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