2
$\begingroup$

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 $\ker T'=\{0_V\}=\{0_U\}$?

$\endgroup$
  • $\begingroup$ Yes this is correct $\endgroup$ – marwalix Feb 1 '15 at 9:11
  • 1
    $\begingroup$ 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
1
$\begingroup$

If $x\in \ker T'$ then

$$T'(x)=T(x)=0$$

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

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

so $x=0_U$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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