Bilinear form/Linear forms/Nondegenerate/Fact/Proof

Proof

(1) follows immediately from bilinearity.
(2). Let ${}u_{1},u_{2}\in V$ and ${}a_{1},a_{2}\in K$. Then, for every vector ${}v\in V$, we have

${}\left\langle a_{1}u_{1}+a_{2}u_{2},v\right\rangle =a_{1}\left\langle u_{1},v\right\rangle +a_{2}\left\langle u_{2},v\right\rangle \,,$

and this means the linearity of the assignment.
(3). Since the assignment is linear by part (2), we have to show that its kernel is not trivial. So let ${}u\in V$ be such that ${}\left\langle u,-\right\rangle$ is the zero mapping. This means that ${}\left\langle u,v\right\rangle =0$ for all ${}v\in V$. This implies, by the definition of nondegenerate, that ${}u=0$.
If ${}V$ has finite dimension, then we have an injective linear mapping between vector spaces of the same dimension. Such a mapping is also bijective, by fact.

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