Proving that Injectivity of a Linear Map is equivalent to Ker(f) = {0}

Question

So, the following is the theorem I'm trying to prove.

Let $f:V \to W$ be a linear map. Then, $f$ is injective iff $Ker(f) = \{0\}$.

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My Proof Attempt:

Let $f: V \to W$ be an injective linear map. Then, we have to show that $Ker(f) = \{0\}$. Clearly, $\{0\} \subset Ker(f)$ since $Ker(f)$ is a subspace of $V$. So, we have to simply show that $Ker(f) \subset \{0\}$.

Let $v \in Ker(f)$. Then, $f(v) = 0$. Now, either $v = 0$ or $v \neq 0$.

If $v = 0$, then we're done. This proves that $Ker(f) = \{0\}$.

If $v \neq 0$, then $f(v) = 0 = f(0)$. So:

$f(v) = f(0) \implies v = 0$

That is a contradiction. Hence, $v = 0$ and $Ker(f) = \{0\}$.

Now, let $Ker(f) = \{0\}$. Then, we have to show that $f$ is injective. Let $x,y \in V$ such that $f(x) = f(y)$. Then:

$f(x) = f(y) \implies f(x)-f(y) = 0 \implies f(x-y) = 0 \implies x-y \in Ker(f)$

Hence, $x-y = 0$ and that proves that $x = y$. That is exactly the condition for injectivity and it proves the desired result.

Can someone look through the argument above and tell me if it is correct or not? Also, I would want to hear any sort of criticism on the quality of my proof-writing.

Answer 1

Your proof is completely correct, but for example the contradiction part is not necessary. If $f(v) = 0$, then you can automatically conclude that $v=0$ since $f(0)=0$ by any linear map and thus by injectivity $v=0$.

Here is how I would write down the proof:

Assume $f$ is injective. Then

$$v \in \ker f \implies f(v) = 0 = f(0) \implies v = 0$$

and thus $\ker f = \{0\}$.

Conversely, assume that $\ker f = \{0\}$ and assume that $f(x) = f(y)$ for $x,y \in V$. Then $$0 = f(x)-f(y) = f(x-y) \implies x-y \in \ker f = \{0\}\implies x-y = 0 \implies x = y$$

demonstrating injectivity. $\quad \square$