A problem on a linear map from a vector space $V$ to the underlying field $\mathbb{F}$

Question

The problem.

Let $T:V \rightarrow \mathbb{F}$ be a linear map. $null\{T\}=\{v \in V : Tv=0_{\mathbb{F}}\}$. Let $u \in V$ such that $u \notin null\{T\}$. To show that $$V=null\{T\} \oplus \{au:a \in \mathbb{F}\}$$

My approach.

First let us show that $V \supset null\{T\} \oplus \{au:a \in \mathbb{F}\}$. Since both $null\{T\}$ and $\{au:a \in \mathbb{F}\}$ are subspaces of $V$, the claim follows from the definition of vector space.

Now let us try to show that $V \subset null\{T\} \oplus \{au:a \in \mathbb{F}\}$. I want to show that given $v \in V$, I can decompose it as $v=u_1+u_2$ such that $u_1 \in null\{T\}$ and $u_2 \in \{au:a \in \mathbb{F}\}$. Applying $T$ to both sides and using linearity, we have $Tv=Tu_2$. Since $u_2 \in \{au:a \in \mathbb{F}\}$, $\exists a^* \in F$ such that $u_2=a^*u$. Hence $Tv=T(a^*u)=a^*Tu$. So given $v \in V$, I need to find $a^* \in \mathbb{F}$ such that $Tv=a^*Tu$. If $F=\mathbb{R}$ or any one-dimensional field, sure. But can I always do this?

Any help would be greatly appreciated.

Answer 1

Hint: Note that for any $v$, $v - \frac{T(v)}{T(u)}u$ is in $null(T)$.

Answer 2

There is, of course a geometric argumentation or you can also compute explicitly the projectors attached to the direct sum.

Solving, for $y\in V$ and $t\in \mathbb{F}$ $$ y=x+t.u\ ;\ x\in null(T) $$ Then one gets $t=\dfrac{T(y)}{T(u)}$ (note that $T(u)\not=0$).

So the two projectors are $$\phi_1(y)= \dfrac{T(y)}{T(u)}.u$$ and $$\phi_2(y)= y-\dfrac{T(y)}{T(u)}.u$$

Check that these are projectors (only one test is needed) and that they have the correct images.