Let $X$ be a square symmetric positive definite matrix.
Show that, $\forall x,y\in\mathbb{R}^n$:
$(a^TXb)^2 \leq (a^TXa)(b^TXb)$ with equality holding iff $a$ and $b$ are linearly dependent.
I'm struggling with this one! Please help.
For equality:
$a$, $b$ linearly dependent $\Leftrightarrow a = kb$.
$\therefore$ on LHS, $(a^TXb)^2 = (a^TXb)(a^TXb)=(kb^TXb)(kb^TXb)=k^2(b^TXb)(b^TXb)$
and on RHS: $(a^TXa)(b^TXb)=(kb^TX(kb))(b^TXb)=k^2(b^TXb)(b^TXb)$ = LHS.
++++++
Question: for a positive definite matrix, X, is it true that, $a^TXa = ||X||.||a||^2$, where ||X|| is the 2-norm?
If so, then I would like to do the following:
$(a^TXb)^2 = |a.Xb|^2 \leq ||a||^2.||Xb||^2 = ||a||^2.||X||^2.||b||^2 = ||a^TXa||.||b^TXb|| = (a^TXa)(b^TXb)$
with equality iff a and Xb are linearly dependent.
my only problem then would be to connect this somehow to the fact that a and b are linearly dependent..? Can someone please tell me if this is completely wrong?