Are morphisms of vector spaces equipped with nondegenerate bilinear forms injective?

Question

A morphism of vector spaces equipped with bilinear forms is a linear map $T: (V, B) \to (V', B')$ such that $B'(Tv, Tw)=B(v,w)$ for all $v, w \in V$.

In the case where the associated quadratic forms to $B$ and $B'$ are positive-definite, then it is certainly the case, since $B'(Tv, Tv) = B(v,v)$ so $Tv=0$ if and only if $v=0$.

But if $B$ and $B'$ are for instance Lorentzian metrics, then this proof no longer works, because we have null vectors.

Can we still prove that morphisms of vector spaces equipped with nondegenerate bilinear forms must be injective? If not, does the additional assumption of symmetry allow us to conclude this?

Also, does anyone know of a good reference that discusses these sort of properties of bilinear forms? Either in the context of modules over rings or vector spaces over fields.

Answer 1

Yes, this is true. Take a basis $e_1,\ldots,e_n$ of $V$. The matrix $(B(e_i,e_j))$ is nonsingular, and equals $(B'(Te_i,Te_j))$, so that is nonsingular. Any linear dependence between the $Te_i$ would make $(B'(Te_i,Te_j))$ singular.

Answer 2

Figured out a coordinate-free proof:

Let $\overline{B}: V \to V^\vee$ and $\overline{B'}: V' \to V'^\vee$ denote the curried mappings corresponding to the bilinear forms (both curried in the same argument). Then it is easy to see by drawing out the commutative diagram that $f$ being a morphism of vector spaces equipped with bilinear forms is equivalent to the requirement that $\overline{B} = f^\vee \overline{B'} f$.

Now if $\overline{B}$ is injective (that is, $B$ is nondegenerate) then by the basic properties of injective functions, $f$ is forced to be injective.

Thus, in fact we do not need to assume $B'$ is nondegenerate to get the desired conclusion.