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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?

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1 Answer 1

up vote 2 down vote accepted

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!

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I prefer the basis line of reasoning, which gives $T(\lambda)=\lambda T(1)=0,$ which seems simpler and more straightforward to me. –  anon Mar 22 '12 at 15:29
Yes, I prefer that to both arguments, but as it's similarly calculational I didn't mention it! –  Matt Pressland Mar 22 '12 at 15:31

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