# Show that a bilinear form is an inner product

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?

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jgens
Gold Member
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 $g(v,w)=v^1w^1 + \dots + v^nw^n$?

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).

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.

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quasar987