# Rank Nullity Theorem application

Show that {T $\in$ $L(R^5, R^4)$: dim null T > 2} is not a subspace of $L(R^5, R^4)$

I have no idea how to show this isn't a subspace the farthest I have gotten is to show that

dim range T < 3

But how would I go from there?

-
Consider $(x_1, x_2, x_3, x_4, x_5)\mapsto (x_1, x_2, 0,0)$ and $(x_1, x_2, x_3, x_4, x_5)\mapsto (0, 0, x_3, x_4)$. –  Git Gud Jul 23 '14 at 1:35

Hint: what does it mean for a set to be a subspace of $L(R^5,R^4)$? How can a subset fail to be a subspace?
Note that $$\pmatrix{ 1&0&0&0&0\\ 0&1&0&0&0\\ 0&0&0&0&0\\ 0&0&0&0&0 } + \pmatrix{ 0&0&0&0&0\\ 0&0&0&0&0\\ 0&0&1&0&0\\ 0&0&0&1&0 } = \pmatrix{ 1&0&0&0&0\\ 0&1&0&0&0\\ 0&0&1&0&0\\ 0&0&0&1&0 }$$