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Verbatim from my textbook

DEFINITION

Let A be a symmetric positive-definite $n × n$ matrix. For two n-vectors v and w, define the A-inner product: $$(v,w)_A = v^TAw $$ The vectors v and w are A-conjugate if $(v,w)_A = 0$.

Note that the new inner product inherits the properties of symmetry, linearity, and positive-definiteness from the matrix A. Because A is symmetric, so is the A-inner product:

$(v,w)_A = v^TAw = (v^TAw)^T = w^TAv = (w,v)_A$. The A-inner product is also linear, and positive-definiteness follows from the fact that if A is positive-definite, then $$(v,v)_A = v^TAv > 0$$ if $v \neq 0$.

I get the symmetry, but why is it that $(v,w)_A$ is positive definite because $(v,v)_A$ is positive definite? How does that follow?

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    Check again how the psd condition is formulated in the definition of scalar products. – 2017-02-20

1 Answers 1

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It's impossible for an inner product to "always be positive", as shown below for your specific case:

Suppose $\langle u, v \rangle > 0$.

Then $\langle u, -v\rangle = u^TA(-v) = -u^TAv = -\langle u, v\rangle < 0$

In the context of inner products, the property that $\langle v, v\rangle > 0$ for $v\neq0$ is referred to as posite-definiteness (and it has nothing to do with $\langle v, w\rangle$).