Show that a bilinear form is an inner product

  • Thread starter CasinelliG
  • Start date
  • #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:

Answers and Replies

  • #2
jgens
Gold Member
1,581
50
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]?
 
  • #3
111
0
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).
 
  • #4
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).
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?
 
  • #5
Bacle2
Science Advisor
1,089
10
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:
  • #6
quasar987
Science Advisor
Homework Helper
Gold Member
4,780
12
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.
 
  • #7
111
0
How does this correlate to the existence of the inverse matrix?
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.
 

Related Threads on Show that a bilinear form is an inner product

  • Last Post
Replies
4
Views
6K
Replies
2
Views
2K
Replies
4
Views
2K
Replies
4
Views
16K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
10
Views
3K
Replies
4
Views
3K
Replies
15
Views
3K
  • Last Post
Replies
3
Views
3K
Top