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I am looking for a proof using the min-max principle. Wikipedia seem to provide just that: http://en.wikipedia.org/wiki/Min-max_theorem#Cauchy_interlacing_theorem

But this part seems to be wrong:

This can be proven using the min-max principle. Let $\beta_i$ have corresponding eigenvector $b_i$ and $S_j$ be the $j$ dimensional subspace $S_j=\operatorname{span}\{b_1,\dots, b_j\}$, then $$ \beta_j = \max_{x\in S_j,\|x\|=1}(Bx,x) =\max_{x\in S_j,\|x\|=1}(PAPx,x) =\max_{x\in S_j,\|x\|=1}(Ax,x)$$

How is the shift from $PAPx$ to $Ax$ legal? $PAP$ is an $m\times m$ matrix while $A$ is an $n\times n$ matrix. $x$ can't fit both. Can anyone correct the proof?

  • 2
    Hint: I am sure you know that $P$ is an orthogonal projection onto $S_j$. So observe that if $x\in S_j$ then $Px=x$.2012-01-01
  • 0
    But P is mxn, how can it preserve x?2012-01-01
  • 0
    Do you know what a projection is?2012-01-01
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    I know that if you project from 3 dimensions to 2 dimensions, it is impossible to have Px=x. Or am I missing something fundamental here?2012-01-01
  • 0
    ,If $P$ is $m\times n$ and $A$ is $n\times n$, how do you find $PAP$?2012-01-01
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    That's right. I believe this is another mistake in this wikipedia article. B=PAP and A are of different dimensions, so P cannot be a square matrix. I think they're missing a transpose on one of the P's2012-01-01

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