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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.

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

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    $\begingroup$ Seems to me that the OP is on a way more basic level than what is required for this to be an answer. $\endgroup$ – Riccardo Sven Risuleo Jun 19 '19 at 6:45
  • $\begingroup$ ;) thanks for the response! $\endgroup$ – Frank Swanton Jun 19 '19 at 7:11

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