Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

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!

share|improve this answer
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.