Right and left radicals in a finite dimensional vector space have same dimension.

Question

I think this is easy but I cannot find how to prove the following:

Consider a vector space of finite dimension $V$ over a field $k$. Let

$$B : V \times V \longrightarrow k$$

be a $k$-bilinear form, $L:= \{ v \in V \mid B(V,v)=0 \}$ and $R:= \{ v \in V \mid B(v,V)=0 \}$ the right and left radicals of $B$.

I want to show that $R$ and $L$ have the same dimension.

Any help would be apreciated.

Answer 1

Write $M$ for the matrix of the form with respect to a basis of $V$, i.e., $m_{i,j}=B(v_i,v_j)$. The dimension of $L$ will be the nullity of $M$ with respect to row vectors, that is $n$ minus the row rank of $M$. Likewise $\dim R$ is $n$ minus the column rank. The row rank equals the column rank.