Let $T\in\mathcal{L} (V)$.

I want to prove that $T/ (\text{null } T)$ (a quotient operator) is injective iff $\text{null } T \cap \text{Range } T=\{0\}$.

I thought of using the property that a linear map is injective iff its null space is $\{0\}$.

In this case , I get $\text{null } (T/\text{null } T) =\{0+\text{null } T\}$ $\iff \{v+\text{null } T:T(v)+\text{null } T=0+\text{null } T\}=\{0+\text{null } T \}$ $\iff \{v+\text{null } T:T(v) \in \text{null } T\}=\{0+\text{null } T \}$, from the definition of affine space.

I'm not sure how to proceed from here.

Maybe we can get $\{v\in \text{null } T : T(v) \in \text{null } T\}=\{0\}$, from using again the definition of affine space, but I don't know how to justify the RHS being just$\{0\}$.

Any help would be appreciated.

  • 1
    $\begingroup$ Can you please remind to me the definition of $T/\ker T$ ? $\endgroup$ Sep 22, 2016 at 20:32
  • $\begingroup$ @mathcounterexamples.net $(T/\ker T)(v+\ker T)=T(v)+\ker T$ $\endgroup$ Sep 22, 2016 at 20:38

2 Answers 2


You are somehow not using your hypothesis.

Assume first that $\text{null } T \cap \text{Range } T=\{0\}$. Then, if $v+\text{null }(T)\in\text{null }(T/\text{null } T)$ this means that $$ Tv+\text{null }T=0+\text{null }T. $$ This implies that $Tv\in\text{null }T$, so $Tv\in\text{null } T \cap \text{Range } T=\{0\}$. That is $Tv=0$, and so $v\in\text{null }T$. In other words, $\text{null }(T/\text{null }T)=\text{null }T$, so $T/\text{null }T$ is injective.

For the converse, assume that $T/\text{null }T$ is injective. Let $y\in\text{null } T \cap \text{Range } T$. Then $y=Tv$ for some $v$, and $$ (T/\text{null }T)(v+\text{null }T)=Tv+\text{null }T=y+\text{null }T=\text{null }T, $$ since $y\in\text{null }T$. As $T/\text{null }T$ is injective, we get that $v+\text{null }T=\text{null }T$, i.e. $v\in\text{null }T$. But then $y=Tv=0$.

  • $\begingroup$ Martin, thanks for your reply. However, in the first direction, I don't understand how you prove injectivity just by saying that $Tv$ is in the intersection of null and range... Also, in the second direction why do you say that «the hypothesis implies (...)» What hypothesis, and why? Thirdly, why do you say that I'm somehow not using my hypothesis? Thanks once again $\endgroup$ Sep 23, 2016 at 20:17
  • $\begingroup$ 1. I added two sentences to the first direction to see if it helps you see it. $$\ $$ 2. We are proving an implication, so the "hypothesis" is what we are assuming. In this case, that $\text{null }(T/\text{null }T)=0+\text{null }T$ ("assume...")$$\ $$ 3. In your question, you only mention the condition $\text{null } T \cap \text{Range } T=\{0\}$ in the statementn of the question and nowhere else, so I thought it was safe to say you were not using it. $\endgroup$ Sep 23, 2016 at 21:13
  • $\begingroup$ Martin, thanks for the edits. I'll read them tomorrow. In response to your 3rd point, I went crazy and tried to prove both directions in one go... $\endgroup$ Sep 23, 2016 at 21:40
  • $\begingroup$ Martin, I still don't see, in your 2nd direction, how the hypothesis allows you to state that... could you please elaborate a bit more. sorry for any inconvenience. $\endgroup$ Sep 24, 2016 at 5:31
  • 1
    $\begingroup$ Yes, the second part was not clear or even wrong. Please see the edit. $\endgroup$ Sep 24, 2016 at 6:30

My last intuition was wrong.

The deduction should be the following: $ \{v+\text{null } T:T(v) \in \text{null } T\}=\{0+\text{null } T \} $ $\iff [ v+\text{null } T=0+\text{null } T \iff T(v) \in \text{null } T]$ $\iff [v\in \text{null } T \iff T(v) \in \text{null } T]$ $\iff [Tv=0\iff Tv \in \text{null } T]\iff [\text{null } T \cap \text{Range } T=\{0\}]$.

We've proved both directions in one go.


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