Prove $\operatorname{Rank} (T) +\operatorname{Nullity} (T) = \dim V$ 
Suppose that $T : V \to W$ is a linear transformation from a finite dimensional vector space $V$ to a vector space $W$. Then $\operatorname{rank} (T)$ is finite and $\operatorname{Rank} (T) +\operatorname{Nullity} (T) = \dim V$.

Here's my proof so far:

Let $n = \dim V$ and $r = \operatorname{rank}(T)$, so we must prove $r = n - \dim \ker(T)$. suppose first that $1 \le r \le n-1$. Let $S = (v_1,\ldots,v_m)$ be a basis for $\ker T$, and choose vectors $T = (v_{m+1},\ldots,v_n)$ in $V$ so that $S \cup T = (v_1,\ldots,v_n)$ is a basis for $V$. we now claim that $U = (Tv_{m+1},\ldots,Tv_n)$ is a basis for range $T$, and with this done we will have $r = \operatorname{rank} T = \dim \operatorname{range} T = \#U = n-m = n - \dim \ker T$.

And from here, I have no idea what to do. My classmate told me that I have to show $U$ is linearly independent and that $\operatorname{Span} U = \operatorname{range} T$ (?) but I'm not quite sure$\ldots$
Am I approaching this theorem too complicatedly? Any help would be very helpful!!

Take $w\in T(V)$. Then $w=T(\alpha_1v_1+\alpha_2v_2+\dots +\alpha_nv_n)$, since $T$ is linear this is equal to:
\begin{align}
\alpha_1T(v_1)+&\alpha_2 T(v_2)+\dots +\alpha_nT(v_n)\\
&=\underbrace{0+0+\dots +0}_m + \alpha_{m+1}T(v_{m+1})+\alpha_{m+2} T(v_{m+2})+\dots+ \alpha_nT(v_n)\\
&=\alpha_{m+1}T(v_{m+1})+\alpha_{m+2} T(v_{m+2})+\dots+ \alpha_nT(v_n).
\end{align}
So $\{T(v_{m+1}),T(v_{m+2}),\dots, T(v_n)\}$ generates $T(V)$.
Now, suppose $\alpha_{m+1}T(v_{m+1})+\alpha_{m+2} T(v_{m+2})+\dots +\alpha_nT(v_n)=0$.
Notice $\alpha_{m+1}T(v_{m+1})+\alpha_{m+2} T(v_{m+2})+\dots +\alpha_nT(v_n)=T(\alpha_{m+1}v_{m+1}+\alpha_{m+2}v_{m+2}+\dots +\alpha_nv_n)\implies \alpha_{m+1}v_{m+1}+\alpha_{m+2}v_{m+2}+\dots +\alpha_nv_n\in \ker(T)$. Therefore $-(\alpha_{m+1} v_{m+1} + \alpha_{m+2} v_{m+2} + \dots +\alpha_nv_n)\in \ker(T)$. So we can write it as $\alpha_1v_1+\alpha_2v_2+\dots +\alpha_mv_m$.
Which leads us to $\alpha_1v_1+\alpha_2v_2+\dots+ \alpha_nv_n=0$, since $\{v_1, v_2,\dots, v_n\}$ is basis for $V$ we conclude $\alpha_i=0$ for all $ 1\leq i \leq n$. in particular $\alpha_{m+1},\alpha_{m+2},\dots, \alpha_n=0$. So $\{ T(v_{m+1}), T(v_{m+2}),\dots, T(v_n)\}$ is linearly independent
