Let $U,V$ be two finite dimensional vector spaces of a field $K$ and let $f:U \times V \to K$a bilinear form. The set $U_0 = \{u \in U: f(u,v) = 0,\forall v \in V \}$ is called the left kernel of $f$ and $V_0 = \{v \in V: f(u,v) =0, \forall u \in U\}$ is called the right kernel of $f$.
Show that $\dim U/U_0 = \dim V/V_0$ and the kernels of $g: U/U_0 \times V/V_0 \to K$ given by $g(u+U_0,v+V_0) = f(u,v)$ is the null space.
The second statement is obvious, but how can I show the first?? that is, $\dim U/U_0 = \dim V/V_0$.
Using the bilinear form, you get a linear map $U/U_0 \to (V/V_0)^*$ and this is injective. Now compare dimensions and you get an inequality (the spaces are finite-dimensional, so dual spaces do not change the dimension).