Finding dimensions directly Let $T:\mathbb{V}\rightarrow \mathbb{W}$ be a linear transformation, with $\dim \ker T=\ell$. Let $\overrightarrow v_{1},...,\overrightarrow v_{k} \in \mathbb{V}$ be linearly independent vectors. Prove that dim$~$span$~ \{T(\overrightarrow v_{1}),...,T(\overrightarrow v_{k})\}\ge k-\ell$.
I can't seem to prove this in a direct way. I can take values for the dimension of the Kernel of $T$, for example:
$\dim\ker T=0 \Rightarrow~$span$~\{T(\overrightarrow v_{1}),...,T(\overrightarrow v_{k})\}=k$
$\dim\ker T=1 \Rightarrow~$span$~\{T(\overrightarrow v_{1}),...,T(\overrightarrow v_{k})\}=k$ or span$~\{T(\overrightarrow v_{1}),...,T(\overrightarrow v_{k})\}=k-1$.
In this fashion, I think there should be an easier way though.

Hint (1) The intersection of two vector subspaces of dimensions $p, q$ itself has dimension $\leq \min\{p, q\}$. (2) Apply the Rank-Nullity Theorem to $\langle {\bf v}_1, \ldots {\bf v}_k\rangle$.
