How to prove $A'B(B'B)^{-1}B'A \leq A'A$, where $A$,$B$ are $n\times k$ matrices and $B'B$ is assumed to be positive definite? I don't see why it is a Cauchy-Schwarz inequality.
1
-
$\begingroup$ If $B^TB$ is invertible, the inequality is equivalent (via Schur complement) to $[A\ B]^T[A\ B]$ being positive semidefinite (which is trivial). Here $A$, $B$ can be any (not necessarily square) matrices with the same number of rows. Can't see CS in it either. $\endgroup$– A.Γ.Commented Aug 31, 2015 at 15:41
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
It is equivalent to show that $$A'(I - B(B'B)^{-1}B')A \geq 0$$ Notice that $H = I - B(B'B)^{-1}B'$ is idempotent and symmetric, hence for any $x \in \mathbb{R}^{n}$, $$x'A'(I - B(B'B)^{-1}B')Ax = (HAx)'(HAx) \geq 0.$$ Hence the result follows.
$\begingroup$
$\endgroup$
1
If $B$ is square (as you've indicated), then the result is obvious:
If $B'B$ is invertible, then so is $B$. So, we note that $$ B(B'B)^{-1}B' = BB^{-1}(B')^{-1}B' = I $$ So, this whole inequality is just $$ A'A \leq A'A $$ which is true, but trivial. It seems that there's something missing from your question.
answered Aug 31, 2015 at 15:35
-
$\begingroup$ Thank you @Omnomnomnom. I made a mistake. A, B are not square. I corrected the question. How to prove then? $\endgroup$– TinaCommented Aug 31, 2015 at 17:39