1
$\begingroup$

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.

$\endgroup$
  • 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 '18 at 17:40
  • $\begingroup$ @WillJagy Right! I just realized this after I posted it. $\endgroup$ – MaTheoPhys1994 Dec 27 '18 at 18:15
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.