# 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

• amazing! does your proof proves Span U = range T as well? Dec 19, 2015 at 1:17
• @Allie what is $U$? Dec 19, 2015 at 1:21

Well, you are indeed almost done: a basis is a linear independent generating set, so as your classmate mentioned, you only need to check these two properties for $v_{m+1},\ldots,v_n$.

For linear independence assume $$0 = \sum_{i=m+1}^n T(v_i)\lambda_i = T(\sum_{i=m+1}^n v_i\lambda_i)$$ and see what this would mean for the $v_{m+1},\ldots,v_n$.

For generation, given a typical element $T(v)$ of $\operatorname{range}(T)$, express $v$ with the basis of $V$ and observe what happens to the $v_1,\ldots,v_m$ when $T$ is applied.