Null space of linear maps plus span of vector

Question

Let $T \in \mathcal{L}(V, F)$, and $u \in V$ such that $T(u) \neq 0$. Prove that $V = \text{null}(T) + \text{span}(u)$

$F$ is just $\mathbb{R}$. My first thought was about the fundamental theorem of linear maps, that is dim$V = $dim null $T + $ dim range $T$, since it's of a similar format, but it doesn't seem to apply. I think the key is that it's a linear map to $\mathbb{R}$, since $T$ sends everything to $0$ or not $0$, but then I think what I don't understand is what exactly is the span of a vector in just $\mathbb{R}$?

Answer 1

Assuming that $V$ is an $F$-vector space, if the map $T:V\to F$ maps one element $u$ to a non-zero element $T(u)$, then $T$ is surjective. Now use the "fundamental theorem of linear maps". Remember that in a $1$-dimensional vector space, the span of every non-zero vector is the whole space.

Answer 2

You can use this theorem. The image of your map is all of $\mathbb{R}$, since if $T(\underline{u}) = r$, then you can hit any $s\in\mathbb{R}$ by considering $T(\frac{s}{r}\underline{u}) = s$. The other way you can see this is by noting that the image is a subspace of $\mathbb{R}$ which isn't just $0$, so it's gotta be the whole thing.

So then you know that the dimension of the kernel of $T$ is $n-1$.

P.S. Are you using Axler? He's the main person I know who calls it the fundamental theorem of linear maps, usually it's "Rank-nullity theorem"

Answer 3

For any $v\in V$, $$ v=\left(v-\frac{T(v)}{T(u)}u\right)+\frac{T(v)}{T(u)}u. $$ The left summand is in $\mathrm{null}(T)$ since $$ T\left(v-\frac{T(v)}{T(u)}u\right)=T(v)-\frac{T(v)}{T(u)}T(u)=0. $$ The right summand is clearly in $\mathrm{span}(u)$.