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!
Best Answer
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