# Equivalent Conditions of Nondegenerate Bilinear Forms and the Gram Matrix

One can often find the following theorem describing equivalent conditions for non degenerate bilinear forms.

$$\textbf{Theorem 1}$$: Let $$V$$ be a vector space over the field $$\mathbb{F}$$ equipped with bilinear form $$\beta : V \times V \to \mathbb{F}$$. The following are equivalent:

(1) Let $$\{ e_i \}$$ be a basis of $$V$$. The matrix $$B = || \beta(e_i, e_j) ||$$ is invertible

(2) $$\forall v \in V / \{ 0 \}, \exists u \in V$$ such that that $$\beta(v,u) \neq 0$$.

We then say a bilinear form is nondegenerate if the above conditions hold for $$\beta$$. Examples of such theorem are provided here in $$\textbf{Proposition} \ 3.11$$ and here in $$\textbf{Theorem} \ 3.1$$.

It is my understanding the matrix $$B := || \beta ( e_i, e_j)||$$ in the above theorem is by definition the Gram Matrix. The Gram matrix then satisfies the following theorem.

$$\textbf{Theorem 2}:$$ If $$V$$ is an vector space equipped with an inner product $$\langle \cdot, \cdot \rangle$$. The set of vectors $$\{ v_1, \ldots, v_n \} \in V$$ is linearly independent iff $$det(B_{ij}) \neq 0$$.

The proof of this theorem is shown in this question.

It appears to me there is a contradiction between these theorems. In $$\textbf{Theorem 1}$$ since $$\{ e_i\}$$ is a basis it's also linearly ind. by definition and therefore by $$\textbf{Theorem 2}$$ (and the invertible matrix theorem) the matrix $$B:= || \beta ( e_i, e_j)||$$ is invertible which then would imply every bilinear form is nondegenerate which can't be true. I am thus failing to recognize some important assumptions. Can someone point out to me what information I am failing to recognize? Thank you for any help.

• your Theorem 2 has an inner product. Over the real field, this is defined positive definite, therefore nondegenerate. There is an analogous version for complexes as well. In brief, an inner product is a very special case of a bilinear form – Will Jagy Dec 27 '18 at 17:40
• @WillJagy Right! I just realized this after I posted it. – MaTheoPhys1994 Dec 27 '18 at 18:15

In the immediate aftermath of writing this question I believe I have uncovered the error in my reasoning. In $$\textbf{Theorem 2}$$ it is already assumed the bilinear form $$\langle \cdot, \cdot \rangle$$ is an inner product which implies it is already nondegenerate by definition and therefore satisfies $$\textbf{Theorem 1}$$. Therefore, there is no contradiction.
Furthermore, for an arbitrary bilinear form $$\beta$$, $$B:= || \beta ( e_i, e_j)||$$ is not invertible in general because not every bilinear is an inner product which means we cannot apply $$\textbf{Theorem 2}$$ to the forms in $$\textbf{Theorem 1}$$. This was my main source of error.