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I am looking for a proof using the min-max principle. Wikipedia seem to provide just that:

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?

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Hint: I am sure you know that $P$ is an orthogonal projection onto $S_j$. So observe that if $x\in S_j$ then $Px=x$. – smanoos Jan 1 '12 at 3:08
But P is mxn, how can it preserve x? – Leo Jan 1 '12 at 9:11
Do you know what a projection is? – smanoos Jan 1 '12 at 13:05
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? – Leo Jan 1 '12 at 13:23
,If $P$ is $m\times n$ and $A$ is $n\times n$, how do you find $PAP$? – smanoos Jan 1 '12 at 21:47

The proof on wikipedia has many flaws indeed. The projection $P$ can be expressed with matrix $P=V*V'$ where columns of $V$ are eigenvectors $v_1, v_2, \dots v_m$ associated with $\alpha_1, \dots \alpha_m$ of $A$. So $V$ is $n*m$ matrix. Then you can write $B=V'*A*V$ and $\max (Bx,x) = \max x'*V'*A*V*x = \max (V*x)'*A*(V*x)$. When $\operatorname{norm}(x)=1$ then $\operatorname{norm}(Vx)=1$ as well. This is how I got to the end of the proof.

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