# 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!!

## 2 Answers

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:

$\alpha_1T(v_1)+\alpha_2 T(v_2)+\dots +\alpha_nT(v_n)=\underbrace{0+0+\dots +0}_m + \alpha_{m+1}T(v_1)+\alpha_{m+2} T(v_2)+\dots+ \alpha_nT(v_n)=\alpha_{m+1}T(v_1)+\alpha_{m+2} T(v_2)+\dots+ \alpha_nT(v_n)$.

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 $\{ T(v_1), T(v_2),\dots, T(v_n)\}$ is linearly independent

• amazing! does your proof proves Span U = range T as well? – Allie Dec 19 '15 at 1:17
• @Allie what is $U$? – Jorge Fernández Hidalgo Dec 19 '15 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.