Show that a bilinear form is an inner product

[CasinelliG]

Hi, I have a bilinear form defined as g : ℝⁿxℝⁿ->ℝ by g(v,w) = v¹w¹ + v²w² + ... + v^n-1w^n-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?

 

[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?

 

[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).

 

[CasinelliG]

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?

 

[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.

 

[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.

 

[Alesak]

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.