# Inner product

Q. Let A be a real invertible n * n matrix. Show that $$\left\langle {\mathop x\limits^ \to ,\mathop y\limits^ \to } \right\rangle \equiv \mathop y\limits^ \to A^T A\mathop x\limits^ \to = \left( {A\mathop y\limits^ \to } \right)^T \left( {A\mathop x\limits^ \to } \right)$$ defines an inner product in R^n, where x and y are column vectors in R^n. What happens when A is not invertible? (Note: M^T is the transpose of a matrix M, obtained by intechanging the rows and columns of M).

The first step would be to show that the inner product is symmetric I would say. I think I should get to $$... = \mathop x\limits^ \to A^T A\mathop y\limits^ \to$$ but I don't know how to do get to it. Can someone suggest a method to use? I'm not sure if I need to explicity write down a matrix in this question.

Related Introductory Physics Homework Help News on Phys.org
(Ay)^t * (Ax) can be interpreted as the "usual" inner product of the vectors Ay and Ax...

matt grime
Homework Helper
Let's drop the silly arrows from vectors, eh? vectors are lower case, matrices are upper case, we'll use ascii pseudo tex so that ' means transpose.

So, we want to show that <x,y>=y'A'Ax is a real inner product.

Firstly since <x,y> is a real number, ie a vector in 1-d then it is symmtric, ie <y,x>=<x,y>. Linearity is even easier since we're just multipliying matrices.

AKG
Homework Helper
You want to show that your given product is symmetric, positive definite, and multilinear. To show it's symmetric:

<x, y> = y'A'Ax
<y, x> = x'A'Ay = (x'A'Ay)' = y'A'Ax as required

The above basically says that a real number is like a 1x1 matrix, which is of course a symmetric matrix. I guess this is what you meant matt? To show it's positive definite

<x, x> = x'A'Ax = (Ax)'(Ax)

Ax is just a vector, and (Ax)'(Ax) is just the sum of the squares of the entries of Ax, so clearly <x, x> > 0 with equality iff x = 0. You can prove linearity on your own. Note you only have to prove linearity in one "component" and the fact that it is symmetric guarantees bilinearity. So just prove:

<ax + y, z> = a<x, z> + <y, z>

matt grime