# generalized Cauchy-Schwarz inequality

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.

• 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.
– A.Γ.
Commented Aug 31, 2015 at 15:41

## 2 Answers

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.

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.

• Thank you @Omnomnomnom. I made a mistake. A, B are not square. I corrected the question. How to prove then?
– Tina
Commented Aug 31, 2015 at 17:39