Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Show that a bilinear form is an inner product

  1. Mar 20, 2012 #1
    Hi, I have a bilinear form defined as g : ℝnxℝn->ℝ by g(v,w) = v1w1 + v2w2 + ... + vn-1wn-1 - vnwn

    I have to show that g is an inner product, so I checked that is bilinear and symmetric, but how to show that it's nondegenerate too?
     
    Last edited: Mar 20, 2012
  2. jcsd
  3. Mar 20, 2012 #2

    jgens

    User Avatar
    Gold Member

    The map you have defined is not an inner product. Let [itex]v = (0,0,\dots,0,1)[/itex] and notice that [itex]g(v,v)=-1 < 0[/itex]. Did you mean to write [itex]g(v,w)=v^1w^1 + \dots + v^nw^n[/itex]?
     
  4. Mar 20, 2012 #3
    Some physics texts define inner product as nondegenerate symmetric bilinear form.

    To show this bilinear form is nondegenerate, write down its matrix and show it´s invertible (= nondegenerate).
     
  5. Mar 20, 2012 #4
    Yes, that's exactly the case. Now the book also says that g is nondegenerate if g(v, w) = 0 for all w in V implies v = 0 (in which V is an arbitrary vector space of dimension n ≥ 1).

    How does this correlate to the existence of the inverse matrix?
     
  6. Mar 20, 2012 #5

    Bacle2

    User Avatar
    Science Advisor

    Doesn't the symmetric part imply that the associated matrix (a_ij):=(q(vi,vj)) is symmetric. Then you have a symmetric matrix.

    Symmetric matrices are diagonalizable, and non-degenerate means none of the diagonal entries is zero.
     
    Last edited: Mar 20, 2012
  7. Mar 20, 2012 #6

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Take a nonzero v. Then there is some vi that is nonzero. Consider w:=(0,...,0,1,0,...,0) where the 1 is in the ith place. Then g(v,w) is nonzero in both cases.

    jgens: An inner product g such that g(v,v)>0 for all nonzero v is a particular case of an inner product called a euclidean inner product.
     
  8. Mar 21, 2012 #7
    Bacle2 said it. It can be also said in this way: every symmetric bilinear form can be diagonalized by succesively completing the square(try it!), until you´ve found a new basis for your vector space where the bilinear form is "diagonal", i.e. of the form x1y1 + x2y2 + ... + xkyk. If k is smaller than dimension of your space, you can easily see the bilinear form is degenerate(by example). In this case, it´s associated matrix is not invertible.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Show that a bilinear form is an inner product
Loading...