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Courant fischer min-max theorem

  1. May 7, 2010 #1
    Assuming the corollary of the Courant Fischer Min Max theorem,

    [tex]\sigma _{1}\geq \cdots \geq \sigma _{q} [/tex] are the singular values of a matrix [tex]M \in \textbf{C}^{m \times n}[/tex] in decreasing order, q = min {m,n}

    [tex]\sigma _{k}(M)= min_{S} max \left\{\left\| Mx\right\| _{2}: x\in S, \left\| x\right\| _{2} =1 \right\} [/tex] , where dim(S)= n-k+1, [tex] S \in \textbf{C}^{n}[/tex].

    we want to prove that

    [tex]\sigma _{i+j-1}(A+B)\leq \sigma _{i}(A) + \sigma _{j}(B) [/tex] for all [tex]i,j=1,2,..., q[/tex] and [tex]i+j \leq q[/tex].

    I have tried to do this using i+j-1 instead of k, and A+B instead of M, I am getting

    [tex]\sigma _{i+j-1}(A+B) \leq min_{S} max \left\{ (\left\| Ax\right\| _{2} +\left\| Bx\right\| _{2}): x\in S, \left\| x\right\| _{2} =1 \right\} [/tex] , where dim(S)= n- (i+j-1) +1.
    But from here, I am not sure if I can distribute the min max to get [tex]min max (\left\| Ax\right\| _{2} )+ min max (\left\| Bx\right\| _{2})[/tex], which in that case would give me the desired result since the singular values are in decreasing order.

    For some reason, I am doubting that .
     
    Last edited: May 7, 2010
  2. jcsd
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