Is it true that $T$ is injective $\iff$ $\mathrm{Ker}(T)=\lbrace\vec{0}\rbrace$?

Question

This is related to a previous question that I posted here. The question this time asks whether or not, if $T:\mathbb{V}\to\mathbb{W}$ is linear, then $T$ is injective $\iff$ $\mathrm{Ker}(T)=\lbrace\vec{0}\rbrace$. I've devised a proof of my own regarding why this is true. Proving $\implies$ is trivial, since we know that for $T$ to be linear, $T(\vec{0})=\vec{0}$, and thus $T(\vec{v})=\vec{0}\implies\vec{v}=\vec{0}$, and thus $\mathrm{Ker}(T)=\lbrace\vec{0}\rbrace$. Proving $\impliedby$ is a little more tricky. Suppose that $\vec{v},\vec{w}\in\mathbb{V}$ are such that $\vec{v}\neq\vec{w}$ and $T(\vec{v})=T(\vec{w})$. Then, since $\vec{v}\neq\vec{w}$, we have that $\vec{v}-\vec{w}\neq\vec{0}$, but since $T(\vec{v})=T(\vec{w})$, we have that $T(\vec{v}-\vec{w})=\vec{0}$, which implies that $\vec{v}-\vec{w}\in\mathrm{Ker}(T)$. But since $\mathrm{Ker}(T)=\lbrace\vec{0}\rbrace$, and $\vec{v}-\vec{w}\neq\vec{0}$, we know that this cannot be possible, and thus $T$ must be injective.

I'm looking to establish whether or not this is true, as it's effectively the only fact that I need to confirm to complete the proof of the Kernel-Rank theorem. Any responses are greatly appreciated, thank you.

Answer 1

Well, you proved it so it is true. You could try the following straightforward, short method

$$T\;\text{ is injective}\;\iff \left(Tv=Tu\implies v=u\right)\iff \left(Tv-Tu=0\implies v=u\right)\iff $$

$$\iff \left(T(v-u)=0\implies v-u=0\right)\iff\left(Tx=0\implies x=0\right)\iff \ker T=\{0\}$$

Answer 2

Yes it is almost OK.

If $T$ is linear then $\{\vec{0}\}\subset Ker (T)$ and not $Ker (T)=\{\vec{0}\}$.

It is equal because $T$ is injective. Take any $\vec{v} \in Ker (T)$. Then $T(\vec{0})= \vec{0}= T(\vec{v})$ we get $\vec{0}=\vec{v}$ and thus $Ker (T)=\{\vec{0}\}$.

Answer 3

$T(u)=T(v)\implies u-v\in kerT$ since the kernel consists of only the zero vector, $u=v$. That's all you need to say for this particular direction.

Answer 4

On the other hand, you can use the rank nullity theorem to prove it. For the kernel is trivial iff the nullity is zero iff $T$ has full rank.