Let $T:V\rightarrow W$ Prove $rg(T)\leq dim(W)$.

Question

Let $T:V\rightarrow W$ Prove $rg(T)\leq dim(W)$.

I try this:

Suppose $dim(V)=n$, $dim(W)=m$ and $B=\{v_1,...,v_n\}$ a basis for $V$.
Let $v\in V$ then exists $a_1 ...a_n\in \mathbb{R}$ such that $v=a_1v_1+...+a_nv_n$ this implies $T(v)=a_1T(v_1)+...+a_nT(v_n)$

Then, $rg(T)=span(T(v_1),...,T(v_n))$.
In this step i'm a little stuck.

If $n>m$ then $rg(T)=dim(W)$

If $m>n$ then $rg(T)=n$ and $rg(T)<dim(W)$ but i don't sure of this. Can someone help me?

Answer 1

Just observe that $\DeclareMathOperator{\span}{span}\span(T(v_1),\dots, T(v_n))$ is a subspace of $W$, hence $$\operatorname{rk}T=\dim\bigl(\span(T(v_1),\dots, T(v_n)\bigr)\le\dim W.$$