# Proving $\dim E=\dim \ker f + \dim \operatorname*{Im}{f}$

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!

• 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. – Dietrich Burde Dec 18 '17 at 15:31
• The basis of E could obtained by extension of the basis of Ker f. You don't need to pick a different basis. – xbh Dec 18 '17 at 15:35

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