There is a linear map $T:V\rightarrow W$, and $U\subseteq V$ is a subspace such that

$$U\cap \ker T=\{0_V\}.$$

I want to prove that the map $T'=T\big|_U:U\rightarrow W$ is also injective. I want to use the fact that a linear transformation, $S:V\to W$ is injective if and only if $\ker S=\{0_V\}$. However, it remains to show that the new function $T'$ is has the property that $\ker T'=\{0_U\}$. Is it sufficient to say that since $U\cap \ker T=\{0_V\}$ we have $ 0_V\in U$, and so no other vector in $U$ is mapped to the zero vector, and thus $\ker T'=\{0_V\}=\{0_U\}$?

  • $\begingroup$ Yes this is correct $\endgroup$ – marwalix Feb 1 '15 at 9:11
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    $\begingroup$ We're assuming $\;T'=T\uparrow_U\;$ , right? Then your argument's intention seems to be right, but there's no actual proof, imo. An idea: $$U\cap\ker T'=U\cap\ker T$$ which follows from $\;T'u=Tu\;,\;\;\forall\,u\in U$ $\endgroup$ – Timbuc Feb 1 '15 at 9:16

If $x\in \ker T'$ then


hence $x\in\ker T$. Since obviously $x\in U$, we have that

$$x\in\ker T\cap U=\{0_V\}=\{0_U\}$$

so $x=0_U$.


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