Dimension of set of all linear maps that map three elements to zero

[Mr Davis 97]

Homework Statement

Note that V and W are finite dimensional vector spaces.
I know how to do the similar problem when we're looking at all of the linear maps that map ALL elements to zero (in this case you can use the rank-nullity theorem to get a definite answer on the dimension). But I am not sure how to approach this one. It seems that upperbound is (dim V)(dim W) if all the vectors $v_1, v_2, v_3$ are just the zero vectors. But I am not sure about the lower bound.

Relevant Equations

rank-nullity theorem?

 

[Office_Shredder]

What are a and b?

It might help to consider the simpler case where just $v_1$ goes to zero. If you're still not sure, pick V and W to be like, $\mathbb{R}^4$ and $\mathbb{R}^3$ and let $v_1=(1,0,0,0)$ and try that case.

 

[Mr Davis 97]

What are a and b?

It might help to consider the simpler case where just $v_1$ goes to zero.

Sorry, $a = \operatorname{dim} V$ and $b = \operatorname{dim} W$.

And yeah, I actually did that and I think I was able to solve it in that case. Basically, if we're trying to find the dimension of $\{ T\in \mathrm{Hom}(V,W) : T(v_1) = 0 \}$, where $v_1$ is nonzero, then if we define the map $\phi : \mathrm{Hom}(V, W) \to W$ where $\phi (T) = T(v_1)$, we get, using the rank-nullity theorem, and recognizing that the image of $\phi$ is $W$, that $\operatorname{dim} (\ker (\phi)) = ab - b$. If $v_1$ is the zero vector, then we just get that $\operatorname{dim} (\ker (\phi)) = ab$.