TITLE: What does a positive definite matrix have to do with Cauchy-Schwarz Inequality?
QUESTION [9 upvotes]: In my text book, Cauchy-Schwarz Inequality is extended to a positive definite matrix.
But I neither understand what the relationship between Cauchy-Schwarz Inequality and a positive definite matrix nor the sentence underlined in red, I am not a strong linear algebra person, I have just started studying it. 
I'd like post the page here so that I might get a better explanation.
Thank you.

REPLY [12 votes]: As a complement to the other answers, I just wanted to note that reasoning a bit more abstractly you can get this result from the ordinary Cauchy-Schwarz inequality for free, without any computation.
The Cauchy-Schwarz inequality is valid in any inner product space (Euclidean vector space if you are working over the real numbers). In a given an inner product space (such as $\Bbb R^n$ with the standard inner product), what a positive definite symmetric linear operator $B$ (given by a positive definite symmetric matrix in the case of $\Bbb R^n$) gives you is precisely another inner product on the same space, defined by $(x\mid y)_B\overset{\rm def}=(x\mid By)=( Bx\mid y)$. Now the Cauchy-Schwarz inequality holds for this inner product as well:
$$
  (x\mid y)_B^2 \leq (x\mid x)_B ~~(y\mid y)_B
\qquad\text{with equality only if $x \parallel y$}
$$
(the notation $x \parallel y$ here means the vectors $x,y$ are linearly dependent, one being a scalar multiple of the other).
Now taking $\def\b{\mathbf b}x=\b$ and $\def\d{\mathbf d}y=B^{-1}\d$, this becomes
$$
  (\b\mid B^{-1}\d)_B^2 \leq (\b\mid \b)_B ~~(B^{-1}\d\mid B^{-1}\d)_B
\qquad\text{with equality only if $\b \parallel B^{-1}\d$.}
$$
which expanding the definition of $(\mid )_B$ simplifies to
$$
  (\b\mid\d)^2 \leq (\b\mid B\b) ~~(\d\mid B^{-1}\d)
\qquad\text{with equality only if $\b \parallel B^{-1}\d$,}
$$
which is your "extended" Cauchy-Schwarz inequality.