# Linear maps take 0 to 0

Suppose $$T$$ is a linear map from $$V$$ to $$W$$ where $$V$$ and $$W$$ are subspaces of a finite-dimensional vector space over some generic field $$\mathcal{F}$$.

Show $$T(0)=0$$.

Proof:
By additivity of a linear map,
(1) $$T(0)=T(0+0)=T(0)+T(0)$$.
Since $$T(0)\in W$$ and $$W$$ is a subspace, there exists an additive inverse for $$T(0)$$: denote as $$k$$.
Add $$k$$ on both sides of (1), we obtain
$$T(0)=0$$.

Is this proof correct?

Reference: Axler, Sheldon J. $$\textit{Linear Algebra Done Right}$$, New York: Springer, 2015.

• What is 'an additive identity for $T(0)$? Do you mean the additive inverse $-T(0)$? – Kavi Rama Murthy Jun 19 '19 at 6:38
• @KaviRamaMurthy Yes. Thanks for the correction! – Frank Swanton Jun 19 '19 at 6:39
• Your proof is correct. – Kavi Rama Murthy Jun 19 '19 at 6:40
• You can note that vector spaces are abelian groups under addition, and a linear map is nothing more than a homorphism with additional structure. – orientablesurface Jun 19 '19 at 7:29

Well, given two groups $$(G,\cdot,e)$$ and $$(G',\circ,e')$$ (think of the additive groups of vector spaces) and a homomorphism $$\phi:G\rightarrow G'$$, i.e., $$\phi$$ is a mapping with $$\phi(g\cdot h) =\phi(g)\circ \phi(h)$$.
Then $$e'\circ \phi(e) = \phi(e) = \phi(e\cdot e) = \phi(e)\circ \phi(e)$$. By multiplying with $$\phi(e)^{-1}$$ from the right, we obtain $$e'=\phi(e)$$. Done.