Bilinear and linear map, then zero map

Question

If $b:V\times V\rightarrow W$ is both bilinear and linear, then it is the zero map. How do I prove this?

What I tried:

linear: $b((x,y)+(x',y'))=b(x,y)+b(x',y')$ and
bilinear:$b((x,y)+(x',y'))=b(x,y)+b(x',y)+b(x,y')+b(x',y')$.

How do I continue from here?

Answer 1

First, you can derive from $b$ being bilinear that for all $x, y \in V$ $$ b(x, 0) = b(0, y) = 0. $$ This is true, since every linear map maps $0$ to $0$. Together with linearity it follows that $$ b(x,y) = b(x, 0) + b(0, y) = 0 $$