4
$\begingroup$

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!!

$\endgroup$
4
$\begingroup$

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

$\endgroup$
  • $\begingroup$ amazing! does your proof proves Span U = range T as well? $\endgroup$ – Allie Dec 19 '15 at 1:17
  • $\begingroup$ @Allie what is $U$? $\endgroup$ – Jorge Fernández Hidalgo Dec 19 '15 at 1:21
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.