3
$\begingroup$

Let $f:E \rightarrow F$ be a linear transformation.

My attempt to solving this:

Let $\dim E=n$ , $\dim \ker{f}=m$ , $\dim \operatorname*{Im}{f}=p$

We need to prove that $n=m+p$.

Let $\{v_1,\dots,v_n\}$ be a basis for $E$.

Let $\{u_1,\dots,u_m\}$ be a basis for $\ker{f}$.

Let $\{t_1,\dots,t_p\}$ be a basis for $\operatorname*{Im}{f}$.

Let $x_1 \in \ker{f} $ Then $x_1$ can be written as a linear combination of $\{u_1,\dots,u_m\}$.Such that $f(x_1)=0$.

Let $x_2 \in \operatorname*{Im}{f} $ Then $x_2$ can be written as a linear combination of $\{t_1,\dots,t_p\}$.

if $x_1 \in \ker{f} $ Then $x_1 \in E$ Since $\ker{f}\subset E$

Then $x_1$ can be written as a linear combination of $\{v_1,\dots,v_n\}$

$x_1=a_1v_1+\cdots+a_nv_n$

if $x_2 \in \operatorname*{Im}{f}$ then $\exists x' \in E /f(x')=x_2$

$x'$ is a linear combination of $\{v_1,\dots,v_n\}$

$x'=c_1v_1+\cdots+c_nv_n$

$x_2=f(x')=c_1f(v_1)+\cdots+c_nf(v_n)$

I have no idea how to continue.

If someone could point out my mistakes and provide what's the best way to tackle these types of problems i would be grateful!

$\endgroup$
2
  • $\begingroup$ The best way to tackle this, I think, is first to compare with a valid proof, e.g., given here. Usually this helps already a lot. $\endgroup$ – Dietrich Burde Dec 18 '17 at 15:31
  • $\begingroup$ The basis of E could obtained by extension of the basis of Ker f. You don't need to pick a different basis. $\endgroup$ – xbh Dec 18 '17 at 15:35
5
$\begingroup$

Let $\{v_1 \dots v_m\}$ be e basis for $\ker(f)$

Let $\{v_1 \dots v_n\}$ be e basis for $E$

We claim that $A=\{T(v_{m+1}) \dots T(v_n)\}$ is a basis for $\operatorname*{Im}(f)$

Let $v \in \operatorname*{Im}(f)$. Then $v=T(w)$ for $w \in E$. This means that $w$ can be expressed as $$w=\sum_{1}^{n}\alpha_iv_i$$

Then $v=T(w)=T(\sum_{1}^{n}\alpha_iv_i)=\sum_{1}^{n}\alpha_iT(v_i)=\sum_{m+1}^{n}\alpha_iT(v_i)$ since the first $m$ terms are in the kernel.

Hence $v \in \operatorname*{span}\{A\}$.

We now have to show linear independence:

Let $\sum_{m+1}^{n}\alpha_iT(v_i)=0 \Leftrightarrow T(\sum_{m+1}^{n}\alpha_iv_i)=0 \Leftrightarrow \sum_{m+1}^{n}\alpha_iv_i \in \ker(f)$

But as each $v_i$ is not in the kernel, this implies that the $\alpha_i$'s are equal to $0$ which proves the claim

$\endgroup$

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.