We are given: $T: V \rightarrow W$ is a linear transformation and $U$ is a subspace of $V$.
$T(U) = \{T(u) | u \in U\}.$
We are also given $S: U \rightarrow T(U)$ is a linear transformation by $S(u) = T(u)$ for all $u \in U$.
I have already proved that $\ker(S) = U \cap \ker(T)$.
I am having a hard time proving: $\dim(T(U)) = \dim(U) - \dim(U \cap \ker(T))$.
Informally, I have said that by removing all the elements that map the linear transformation $T(u)$ to $\vec{0}$ will give you the dimension of $T(U)$ but I am having a hard time showing this mathematically.
Can someone give me some hints or point me in the right direction to go about doing this?
$dim(T(U)) = dim(S(U))$, so re-write the equation in terms of $S$ as $dim(S(U)) = dim(U) - dim(ker(S))$ and apply the rank-nullity theorem.