Proof that Vector Space in Domain of Linear Map is a Direct Sum

Question

I'm working through problems in Linear Algebra just for fun and I am getting stuck on Axler 3.4.

Suppose that $T$ is a linear map from $V$ to $\mathbf{F}$. Prove that if $u \in V$ is not in $null\ T$, then \begin{align*} V=null\ T \oplus \left\{au:a\in F\right\} \end{align*}

My proof so far:

Suppose that $T$ is a linear map from $V$ to $\mathbf{F}$ and $u \in V$ is not in $null\ T$. We know that $Tu \neq 0$ since $u \notin null\ T$. Since $T$ is a linear map, $T(au)=aTu$. $aT(u)=0$ only when $a = 0$ since $Tu \neq 0$. Therefore, we can show that \begin{align*} null\ T \cap \left\{au: a \in F\right\} = \left\{0\right\} \end{align*} Let $v \in V$ be any vector. We want to show that $V=null\ T + \left\{au: a \in F\right\}$.

This is where I am getting stuck. I have tried using the linearity and homogeneity property of linear maps but I'm not getting anywhere. I would really appreciate a hint on how to proceed and solve this problem!

Answer 1

You have two choices to complete your proof

• the easiest is to notice using the rank nullity theorem that $$\dim V=\dim\ker T+\dim\operatorname{span}(u)$$
• the second way is: let $x\in V$ so $$x=\underbrace{x-\frac{Tx}{Tu}u}_{\in\ker T}+\underbrace{\frac{Tx}{Tu}u}_{\in\operatorname{span}(u)}$$