[Proof]. A linear operator T is injective iff $N(T)= \{0\}$

Question

Part 1

Let $T:X \rightarrow Y$ be a linear operator which is injective, Since $T0=0$ for $0 \in X$. Pick $x \in X$ s.t $x \neq 0$ this would imply that $Tx \neq 0$ (as $T$ is injective) then the set of all vector in $X$ which makes $Tx=0$ is the singelton set $\{0\}$ $\implies$ $N(T)=\{0\}$.

Conversely,

Let $N(T)=\{0\}$ $\implies$ [ $Tx=0 \implies x=0$ for $x \in X$ ] now if we pick $x_o \neq 0 \implies x_o \notin N(T) \implies Tx_o \neq T0 \implies$ $T$ is injective.

Answer 1

You do need linearity somewhere.

If $T$ is injective, then $T(0)=0$ and so $x \neq 0$ implies $T(x) \neq T(0)=0$. This means that $N(T)=\{0\}$.

Suppose that $N(T)=\{0\}$, then suppose $T(x) = T(y)$. Then $0=T(x)-T(y) = T(x-y)$ by linearity and so $x-y = 0$, as $x-y \in N(T)=\{0\}$. Hence $x=y$ and $T$ is indeed injective.