Nullity of composition of linear maps Let $T : \mathbb{R}^n → \mathbb{R}^k$ and $U : \mathbb{R}^k → \mathbb{R}^m$ be linear maps. How can I show that 
nullity(U ◦ T) ≤ nullity(U) + nullity(T)?
I am certain that I will have to use the rank-nullity theorem but I do not know how to use it and prove the above statement.
Any solution without using restrictions?
 A: The proof below, if correct, is based in a slightly general case. I solved this exercise some time ago (not very much time ago) and I had in my notes (this is the exercise 3.B.22 of Linear algebra done right third edition).
About notation: FTLA is the named fundamental theorem of linear algebra that state that
$$\dim V=\dim T(U)+\dim(\ker T)$$
for $T\in\mathcal L(U,V)$.

Let $U,V$ finite-dimensional vector spaces and $S\in\mathcal L(V,W)$, $T\in\mathcal L (U,V)$, and we must show that 
$$\dim\ker(ST)\le\dim\ker(T)+\dim\ker(S)\tag{1}$$
Now because $U':=\ker(ST)$ is a subspace of $U$ with $T':=T|_{U'}$ applying the FTLA we have that
$$\dim U'=\dim\ker(T')+\dim T'(U')\tag{2}$$
and because $\ker(T')=\ker(T)$ then $(2)$ becomes
$$\dim\ker(ST)=\dim\ker(T)+\dim T'(U')\tag{3}$$
Now if $T$ is surjective $\dim T'(U')=\dim\ker(S)$ so $(3)$ becomes the equality in $(1)$.
If $T$ is not surjective is possible that $\dim T'(U')<\dim\ker(S)$, when exists some $Sv=0$ such that $v\notin T(U)$. In any case we have that
$$T'(U')=T(U')\subseteq \ker(S)$$
because if $STu=0$ then $Tu\in\ker(S)$. Hence
$$\dim T'(U')\le\dim\ker(S)$$
and $(1)$ holds.

Ah, I had seen that the rank nullity theorem is equivalent to the FTLA. Sorry, I dont know a way to solve this exercise without some version of this theorem.
