= Solution
A <bilinear form> $B$ on $V$ is <nondegenerate bilinear form>[nondegenerate] when
$$
B(v,w)=0\quad\text{for every }v\in V
\qquad\Longrightarrow\qquad w=0.
$$
In finite dimensions this is equivalent to nondegeneracy in the first argument.
Define the <linear map>
$$
\beta_1:V\longrightarrow V^*,
\qquad
\beta_1(w)=B_1(-,w).
$$
Nondegeneracy makes $\beta_1$ injective. Since $V$ and its <dual space> have the same finite dimension, $\beta_1$ is an isomorphism. Similarly define $\beta_2(w)=B_2(-,w)$ and set
$$
\boxed{\alpha=\beta_1^{-1}\beta_2}.
$$
Then $\alpha$ is linear and
$$
B_1(v,\alpha w)=\beta_1(\alpha w)(v)
=\beta_2(w)(v)=B_2(v,w).
$$
If $w\in\ker\alpha$, this identity gives $B_2(v,w)=0$ for every $v$. Conversely, if $B_2(v,w)=0$ for every $v$, then $B_1(v,\alpha w)=0$ for every $v$, so nondegeneracy gives $\alpha w=0$. Hence
$$
\boxed{\{w\in V:B_2(v,w)=0\text{ for every }v\in V\}=\ker\alpha}.
$$
This is the <representation of a bilinear form relative to a nondegenerate bilinear form>.
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