| dbp:proof
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- WLOG, is diagonalized, then we need to show
By the standard Hölder inequality, it suffices to show
By the Schur-Horn inequality, the diagonals of are majorized by the eigenspectrum of , and since the map is symmetric and convex, it is Schur-convex. (en)
- And it remains to find some such that .
If , then any would work. Otherwise, if , then any would work, and so on. If none of these work, then it means , contradiction. (en)
- Part 2 is a corollary of part 1, by using .
By Poincare’s inequality, is an upper bound to the right side.
By setting , the upper bound is achieved. (en)
- The second is the negative of the first. The first is by Wielandt minimax. (en)
- The case.
Let , and any , it remains to show that
To show this, we construct an orthonormal set of vectors such that . Then
Since , we pick any unit . Next, since , we pick any unit that is perpendicular to , and so on.
The case.
For any such sequence of subspaces , we must find some such that
Now we prove this by induction.
The case is the Courant-Fischer theorem. Assume now .
If , then we can apply induction. Let . We construct a partial flag within from the intersection of with .
We begin by picking a -dimensional subspace , which exists by counting dimensions. This has codimension within .
Then we go down by one space, to pick a -dimensional subspace . This still exists. Etc. Now since , apply the induction hypothesis, there exists some such that Now is the -th eigenvalue of orthogonally projected down to . By Cauchy interlacing theorem, . Since , we’re done.
If , then we perform a similar construction. Let . If , then we can induct. Otherwise, we construct a partial flag sequence By induction, there exists some , such that thus (en)
- Part 2 is a corollary, using .
is a dimensional subspace, so if we pick any list of vectors, their span must intersect on at least a single line.
Take unit . That’s what we need.
: , since .
: Since , we find . (en)
- Let S' be the closure of the linear span .
The subspace S' has codimension k − 1. By the same dimension count argument as in the matrix case, S' ∩ Sk has positive dimension. So there exists x ∈ S' ∩ Sk with . Since it is an element of S' , such an x necessarily satisfy
:
Therefore, for all Sk
:
But is compact, therefore the function f = is weakly continuous. Furthermore, any bounded set in H is weakly compact. This lets us replace the infimum by minimum:
:
So
:
Because equality is achieved when ,
:
This is the first part of min-max theorem for compact self-adjoint operators.
Analogously, consider now a -dimensional subspace S'k−1, whose the orthogonal complement is denoted by S'k−1⊥. If S' = span{u1...uk},
:
So
:
This implies
:
where the compactness of A was applied. Index the above by the collection of k-1-dimensional subspaces gives
:
Pick S'k−1 = span{u1, ..., u'k−1} and we deduce
: (en)
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