3
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ 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 $\endgroup$
    – Will Jagy
    Dec 27, 2018 at 17:40
  • $\begingroup$ @WillJagy Right! I just realized this after I posted it. $\endgroup$ Dec 27, 2018 at 18:15

1 Answer 1

1
$\begingroup$

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.

In summary, every inner product is a bilinear form but every bilinear form is not an inner product as a result of their respective definitions.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .