Prove injectivity is equivalent to null space equals $\{0\}$

Question

I think I got the intuition of the reasoning. Because injectivity means one-to-one, if null space is larger than $\{0\}$, there are more than one way to transform $V$ to $\{0\}$. So it is not injective.

But I don't understand the proof.

Proof:

First suppose $T$ is injective. We want to prove that null $T = \{0\}$. We already know that $\{0\} \subset \text{null }T$ (I understand). To prove the inclusion in the other direction (what's that mean?), suppose $v \in \text{null } T$. Then $$T(v) = 0 = T(0)$$ Because $T$ is injective, the equation above implies that $v=0$. Thus we can conclude that $\text{null }T = \{0\}$, as desired.

To prove the implication in the other direction (what's that mean?), now suppose $\text{null }T = \{0\}$. We want to prove that $T$ is injective. To do this, suppose $u, v \in V$ and $Tu = Tv$. Then $$0 = Tu - Tv = T(u-v)$$

is this step just coming from $Tu = Tv$? $Tu - Tv = 0$

Thus $u-v$ is in null $T$, which equals to $\{0\}$. Hence $u-v = 0$, which implies that $u=v$. Hence $T$ is injective, as desired

Answer 1

The statement is$$T\text{ is injective}\iff\operatorname{null}T=\{0\}.$$So, there are actually two things to prove:

1. $T\text{ is injective}\implies\operatorname{null}T=\{0\}$;
2. $\operatorname{null}T=\{0\}\implies T\text{ is injective}.$

Now, in order to prove the first statement, we have to prove that if $T$ is injective, then the sets $\{0\}$ and $\operatorname{null}T$ are equal. But, since it is trivial that $\{0\}\subset\operatorname{null}T$, what remains to be proved is the opposite inclusion, which is $\operatorname{null}T\subset\{0\}$.

The second statement is the implication in the other direction. Finally, the equality $T(u)-T(v)=T(u-v)$ comes from the fact that the map $T$ is linear.

Answer 2

Proving the inclusion in the other direction means proving the reverse inclusion: $\ker T\subset\{0\}$.

Provin the implication in ther direction means this:

The author has just proved that injectivity implies $\ker T=\{0\}$. ($\Rightarrow$). Now he/she going to prove the reverse implication: $$\text{injectivity}\Leftarrow \ker T=\{0\}.$$

Fianlly the step you mention comes from the hypothesis $Tu=Tv\;$ AND $\;$linearity of $T$: $\; Tu-Tv=T(u-v)$.