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: 
The two answers presented above appear to contradict each other - let me resolve that contradiction.

Lob proved that, for all sentences $\varphi$, $$PA\vdash "(PA\vdash\varphi\implies\varphi)"\implies PA\vdash\varphi;$$ if PA proves that, if PA proves $\varphi$ then $\varphi$ is true, then PA proves $\varphi$. Intuitively, any "reasonable" expression of "I am provable" in the language of arithmetic has the property that PA proves that it is true iff it is provable, so any such sentence should be provable.

Meanwhile, say that a formula $\psi$ *expresses provability* if for all sentences $\theta$, $PA\vdash \psi([\theta])$ holds iff $PA\vdash\theta$ (where "$[\cdot]$" is the Goedel number operation).

What Kreisel proved was that there is a formula $\psi$ expressing provability, such that PA does *not* prove "$\psi($me$)$", the fixed point sentence for $\psi$.

The apparent contradiction is resolved by the following:

> 
> Just because $\psi$ expresses provability, doesn't mean that PA *proves* that $\psi$ expresses provability.
> 
> 
> 

Lob's theorem implies that *natural* expressions of provability in PA - ones for which PA proves all the relevant basic properties - the resulting interpretation of "I am provable" is, in fact, provable.
Your sentence is constructed to have the property that
$$ H \leftrightarrow (\mathsf{PA}\vdash H) $$
In particular, then, PA proves
$$ (\mathsf{PA}\vdash H) \to H $$
This is the premise for [Löb's theorem](https://en.wikipedia.org/wiki/L%C3%B6b's_theorem) which then concludes that PA proves $H$ itself.

So $H$ is true!