Let $(V,\mathbb{K})$ be a vector space over some field $\mathbb{K}$ and let $T:\mathbb{K}\rightarrow V$ be a
linear map such that $T(1) = 0$. I am trying to figure out whether it is necessarily the case that $T$ is
the $0$-map, i.e., the map which sends every element of $\mathbb{K}$ to $0 \in V$.
I believe this is true because, since $T(0) = 0$ for any linear map,
$$
0 = T(0) = T(\lambda \cdot 0) = \lambda T(0) = \lambda T(1) = T(\lambda)
$$
where $\lambda \in \mathbb{K}$. Since no special conditions were placed on the selection of $\lambda$,
$T(\lambda) = 0$ for all $\lambda \in \mathbb{K}$. I believe then that this implies that $T$ is the $0$-transformation.
Is this the right way to look at things? I think the calculation/conclusion is correct but I think I might
be missing part of the bigger picture. Is there a theorem, for example, that would make this result immediate
without having to go through the setup/computation above?
Best Answer
One way to think about it is to recall that linear maps are defined by how they act on a basis, and $1$ is a basis of $\mathbb{K}$. In particular, the image of a map is spanned by the images of basis vectors of the domain. In this case, the image of $T$ is spanned by $T(1)=0$, so the image of $T$ is $\{0\}$ and $T$ must be the zero map.
Personally I think I prefer your calculation though!