I tried to solve this problem, but, I can't. I need help.
Let $U$ and $V$ a finite-dimensional vector spaces, $T:U\rightarrow V$ and $S:V\rightarrow W$ linear transformations. Proof that $Nul(S\circ T)\leq Nul(S)+Nul(T)$
($Nul(T)$ means the dimension of the Kernel of $T$ and $Ran(T)$ means the dimension of the image of $T$)
My incomplete proof:
We know that $S\circ T: U\rightarrow W$, then
$Nul(S\circ T)+Ran(S\circ T)=Dim(U)=Nul(T)+Ran(T)$
Now, $Nul(T)+Ran(T)\leq Nul(T)+dim(V)$ because the image of $T$ is a subespace of $V$.
$Nul(T)+dim(V)=Nul(T)+Nul(S)+Ran(S)$.
Hence, $Nul(S\circ T)+Ran(S\circ T)\leq Nul(T)+Nul(S)+Rang(T)$...
But I do not know how to conclude. I think that if $Ran(S\circ T)\geq Ran(T)$, the proof it's ok because in the last inequality we obtain $Nul(S\circ T)\leq Nul(T)+Nul(S)$, but, it is true?