Hello I have the following question :

$W,V$ linear spaces with finite dimension above the same field.

$$T : V \rightarrow W$$ T is a linear transformation, let $U$ span of $V$ that impiles $V=U \oplus kerT$

$$S:U \rightarrow ImT$$

S is a linear transformation that is defined by $S(u)=T(u)$ for all $ u\in U $

Proof linear transformation $S$ isomorphism.

This is what I managed to get, I'm not so sure that I went in the right direction with this proof.


We know that : $$dimV=dimU+dimKerT$$ And as well that $$dimV=dimImT+dimKerT$$

Therefore we can conclude that $dimU=dimImT$

Any Idea how to process from this?

Thank you!

  • $\begingroup$ You're idea for a proof is wrong unfortunately: There is not reason why the kernel of $T$ should be zero. $\endgroup$ – a student Aug 9 '15 at 23:56
  • $\begingroup$ @astudent I tried different idea, I think that could help me inorder to proof that S isomorphism. but not sure how, any ideas? $\endgroup$ – JaVaPG Aug 10 '15 at 0:24

Your proof is excellent so far! You have $S:U \rightarrow \operatorname{Im}T$ is a linear transformation with $\dim(U) = \dim(\operatorname{Im}T)$.

In order to prove that $S$ is an isomorphism, it now is sufficient to show that $\ker(S) = \{0\}$. However, we know that this is true since $V = U \oplus \ker T$, which by definition implies that $U \cap \ker T = \{0\}$.

  • $\begingroup$ Thank you for your answer, just to make sure how you managed to reach to $KerS=\{0\}$ from $U \cap \ker T = \{0\}$ $$$$ we know that $KerS=KerT$ since $S(u)=T(u)$ for all $u \in U$ and as well we know that $ kerS \subseteq U$ therefore $kerS \cap U = \{0\}$ implies that $kerS=\{0\}$ $\endgroup$ – JaVaPG Aug 10 '15 at 9:33
  • $\begingroup$ Yes, that's a reasonable way to put it $\endgroup$ – Omnomnomnom Aug 10 '15 at 11:37
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    $\begingroup$ More concisely: $\ker S \subseteq \ker T$ and $\ker S \subseteq U$ implies that $\ker S \subseteq U\cap\ker T = \{0\}$. So, $\ker S=\{0\}$. $\endgroup$ – Omnomnomnom Aug 10 '15 at 11:42
  • $\begingroup$ All clear, thank you very much! $\endgroup$ – JaVaPG Aug 10 '15 at 11:53
  • $\begingroup$ I'm doing the same question again and looking at it again, I think that the conclusion $\ker S \subseteq \ker T$ is wrong, I can understand that $ker S=ker T$ since we know that for all $u \in U \rightarrow S(u)=T(u)$ but since $\ker S \subseteq U$ and also $U \cap ker T = \{0\}$ Therefore $ker S \cap ker T \{0\}$ I don't understand how you managed to get $ker S \subseteq ker T$ can you please explain? $\endgroup$ – JaVaPG Aug 11 '15 at 20:41

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