Proving Injectivity for a Linear Transformation iff $\ker(T) = \{0\}$

Question

Let $V$ and $W$ be finite dimensional vector spaces over a field $F$. Let $T: V \to W$ be a linear transformation.

For $T$ to be injective, there must not exist two vectors not equal to each other such that

$$T(w) = T(v) \ \ s.t. \ \ w, v \in V$$

I am trying to prove that this implies that $\ker(T) $ must contain nothing but merely the zero vector in $V$. I don't know if my proof is valid though. I'll go through with it now.

Let there be a vector $u, s \in V$ such that $T(u) = T(s)$.

$$\implies T(u) - T(s) = \vec 0$$

Since, $T$ is a linear transformation,

$$\implies T(u-s) = \vec 0$$

If $u \neq s \implies u-s \in \ker (T)$, as $u-s$ and $0$ in $V$ both map to $\vec 0_w$. However, if this is so far okay, I seem to state that $T(\vec 0_V) = \vec 0_w$ without proving it, and I don't seem to know how. If this is indeed correct so far, how do I then prove this?

Answer 1

It depends a bit on the precise definition of linearity that is chosen, but this works most of the time: $$ T(0_V)=T(0_V+0_V)=T(0_V)+T(0_V) $$ Now take $T(0_V)$ away from both sides.

Answer 2

Hint: Let $T(0)=v$. Then $$v=T(0)=T(0+0)=T(0)+T(0)=2v$$ thus $v=?$

Answer 3

For every linear map, it is true that $T(0)=0$. This is because $T(0)=T(0+0)=T(0)+T(0)$, and therefore $0=T(0)-T(0)=\bigl(T(0)+T(0)\bigr)-T(0)=T(0)$.

But in order to prove that your condition implies that $\ker T=\{0\}$, you could do this: if $\ker T$ contains some vector $u\neq0$ such that $T(u)=0$, then $T$ would not be injective, because $u\neq0$, but $T(u)=T(0)=0$.