# Proving dimensions for arbitrary linear transformations

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?

• @YadatiKiran: $T(U) \subset W$ and $U \subset V$, why would the intersection be relevant? – copper.hat Mar 10 at 5:42

$$dim(T(U)) = dim(S(U))$$, so re-write the equation in terms of $$S$$ as $$dim(S(U)) = dim(U) - dim(ker(S))$$ and apply the rank-nullity theorem.