Show that a bilinear form is an inner product

1. Mar 20, 2012

CasinelliG

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. Mar 20, 2012

jgens

The map you have defined is not an inner product. Let $v = (0,0,\dots,0,1)$ and notice that $g(v,v)=-1 < 0$. Did you mean to write $g(v,w)=v^1w^1 + \dots + v^nw^n$?

3. Mar 20, 2012

Alesak

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. Mar 20, 2012

CasinelliG

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. Mar 20, 2012

Bacle2

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
6. Mar 20, 2012

quasar987

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. Mar 21, 2012

Alesak

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.