Let $A:V \rightarrow W$ be a linear map. Prove that A is injective iff $\{v \in V :Av=0\}=\{0\}$ I read that a linear transform is injective iff the kernel of the function is $0$, but I am supposed to use the fact that for all $a,b \in V$, $A(a+b)=A(a)+A(b)$. Is this any different than proving a function is injective by using the implication "If $f(x)=f(y)$,then $x=y$"?
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$\begingroup$It is not any different than proving a function is injective since linear mappings are in fact functions as the name suggests. However linear maps have the restricted linear structure that general functions do not have. It is for this reason that we often consider linear maps as general results are possible; few general results hold for arbitrary maps. This allows us to easily prove injectivity.
Suppose you have that $A$ is injective. Then we want to conclude that the kernel of $A$ is $0$. Suppose $x\in\ker A$, then $A(x) = 0$. However we know that $A(0) = 0$ since $A$ is linear. Since $A$ is injective and $A(x) = A(0)$, we must conclude that $x = 0$.
Can you handle the other direction? Here's a hint: suppose $x,y\in V$ and $Ax = Ay$, then $A(x-y) = 0$ by making use of linearity.
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