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