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Inner product

  1. Sep 9, 2005 #1
    Hi can someone please help me get started on the following question?

    Q. Let A be a real invertible n * n matrix. Show that [tex]\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)[/tex] 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 [tex]... = \mathop x\limits^ \to A^T A\mathop y\limits^ \to [/tex] 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.
     
  2. jcsd
  3. Sep 9, 2005 #2
    (Ay)^t * (Ax) can be interpreted as the "usual" inner product of the vectors Ay and Ax...
     
  4. Sep 9, 2005 #3

    matt grime

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    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.
     
  5. Sep 9, 2005 #4

    AKG

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    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>
     
  6. Sep 9, 2005 #5

    matt grime

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    the "unofficial convention" for scalars is to use letters like r,s,t (preferably greek letters like lambda but i can't do that in plain html) for them, reserving u,v,w,x,y,z for vectors and A,B for matrices.

    and yes, that was what i meant, AKG about 1x1 matrices and real numbers being the same thing.
     
  7. Sep 9, 2005 #6
    Thanks for the help guys. When I write vectors I normally use a tilde, I just used arrows because I didn't know how to put a little squiggle underneath the vectors.
     
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