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Homework Help: Matrix Analysis (Functional Analysis) Question

  1. Jun 1, 2012 #1
    1. The problem statement, all variables and given/known data
    Let [itex] \lambda_1 ,..., \lambda_n [/itex] be the eigenvalues of an [itex] nXn[/itex] self-adjoint matrix A, written in increasing order.
    Show that for any [itex] m \leq n [/itex] one has:
    [itex] \sum_{r=1}^{m} \lambda_r = min \{ tr(L) :dim(L) =m \} [/itex] where [itex] L [/itex] denotes any linear subspace of [itex] \mathbb {C} ^n [/itex], and [itex] tr(L):= \sum_{r=1}^{m} Q( \Phi_r) [/itex] for some orthonormal basis [itex] \{ \Phi _r \} [/itex] of [itex] L [/itex].

    (Q is the quadratic form associated with the inner product).

    2. Relevant equations
    3. The attempt at a solution
    I really have no idea on how to start this.
    On the one hand, I think the trace will always be equal to m, which means I'm probably getting it wrong...

    Hope you'll be able to help me

    Thanks in advance
  2. jcsd
  3. Jun 1, 2012 #2
    I know perhaps next to nothing about functional analysis, but I'd like to try helping.

    I think <phi,phi>=1, while Q(phi)=<A phi,phi>=lambda.

    This is just a guess at the idea, I'm not sure where it would go from there.
  4. Jun 1, 2012 #3
    Then maybe, since the lambdas are written in increasing order, the minimum trace over subspaces will somehow find you the sum of lower m lambdas.
  5. Jun 1, 2012 #4
    I feel the difficulty with this question is more notationalwise. Perhaps it would be a good idea to first try to prove it for m=1. The general idea is very similar.

    So, we want to prove that

    [tex]\sum_{r=1}^m \lambda_r = \min\{tr(L)~\vert~ \dim(L)=m\}[/tex]

    A good first step would be to show the existence of a subspace L with [itex]\dim(L)=m[/itex] such that

    [tex]\sum_{r=1}^m \lambda_r = tr(L)[/tex]

    We know that a hermitian matrix A always has a orthonormal base of eigenvectors. Use this base of eigenvectors to find a suitable L.
  6. Jun 2, 2012 #5
    Thanks a loth to you both!

    Here is my attemp:
    We have a Hermitian matrix A, which implies that we have an orthonormal base of eigenvectors to the entire space. Since we have [itex]n[/itex] eigenvalues, we must have [itex]n[/itex] eigenvectors, and by the assumption, we can choose form such that they'll form an orthonormal basis: [itex] \{ v_1 ,..., v_n \}[/itex] where [itex]Av_i = \lambda_i v_i [/itex] .
    Now my guess was that the space L we need to choose is [itex]span\{v_1,...,v_m \}[/itex].

    If the quadratic form is assumed to be the one defined by "algebrat" , then we'll indeed get the needed equality.
    But how can I prove this is the minimum?

    Thanks a lot again ! (and hope you'll be able to help me finish this)
  7. Jun 2, 2012 #6
    For the minimum case, perhaps it's better to first prove a special case to see what happens. So take m=1 and n=2.

    So take a subspace L of dimension 1. This is generated by one element (of norm) 1. Call this element v.
    Since there is an orthonormal basis of the entire space (call it {x,y}). We can write [itex]v=ax+by[/itex] for scalar a and b.

    Now, what is [itex]<Av,v>[/itex]?? Use that [itex]v=ax+by[/itex].
  8. Jun 2, 2012 #7
    OK. Let's see: We have eigenvalues [itex] \lambda_1 , \lambda_2 [/itex], with corresponding eigenvectors [itex]x,y [/itex].
    We'll get: [itex] <Av,v>= a^2 \lambda_1 + b^2 \lambda_2 [/itex] . What you mean is obviously that this expression is minimal when [itex] a=1,b=0 [/itex] since [itex] \lambda_1 \leq \lambda_2 [/itex].
    But what if we take [itex] b=0[/itex] and a smaller [itex] a [/itex] ?

    Thanks a lot again !
  9. Jun 2, 2012 #8
    Notice that v must have norm 1. So this places conditions on a and b.
  10. Jun 2, 2012 #9
    Great ! It implies that a+b=1 !

    Then we're done!

    Thanks a lot !

    I'll try generlize this idea. If I won't succeed I'll reply here agin

    Thanks again !
  11. Jun 2, 2012 #10
    That [itex]a^2+b^2=1[/itex].

    Anyway, the general case is quite similar.
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