I have the following question with me:
Let $V$, $W$ be finite dimensional vector spaces over F. For a bilinear map $B$ : $V × W$ → $F$, define its left and right radicals by: $Lrad(B)$ = {$v ∈ V : B(v, x) = 0$ for all $x ∈ W$} and $Rrad(B)$ = {$w ∈ W : B(x, w) = 0$ for all $x ∈ V$ }. Prove that $$dim V − dim Lrad(B) = dim W − dim Rrad(B)$$
I know that the first term would be the dimension of the linear function on W, $B(v',.)$. which I can get by considering the linear map which sends the vector $v$ to $B(v',w)$ where $v'$ is the corresponding element in $V/Lrad(B)$. Simiarly I get the RHS also. But I do not know how to show that they are equal.
It would be helpful if someone could help me in finishing the argument.
Let $V_1=V/Lrad(B)$ and $W_1=W/Rrad(B)$. Define $B_1:V_1\times W_1\longrightarrow F$ by $B_1(\overline{v},\overline{w}):=B(v,w)$, for all $v\in V,w\in W$. Note that $B_1$ is a well-defined bilinear map. Also, $Lrad(B_1)=Rrad(B_1)=\{0\}$.
Now $\theta:V_1\longrightarrow W_1^\ast$ defined by $\theta(x)(y)=B_1(x,y)$ for all $x\in V_1,y\in W_1$ is a linear map with $\ker(\theta)=Lrad(B_1)=\{0\}$. Thus $\dim(V_1)\leq \dim(W_1^\ast)=\dim(W_1)$. Similarly $\dim(W_1)\leq \dim (V_1)$. Hence, the proof.