I was given this question in my introductory linear algebra class:
Given a linear transformation $T: V \rightarrow \tilde{V}$, Prove that $T$ is injective iff $\ker(T) = \{0_V\}$.
I came up with the following proof:
Let us assume to the contradictory that $T$ is not injective even though $\ker(T) =\{0_V\}$. Then there exist $u, v \in V$ such that $T(u) = T(v)$ for $u \neq v$. We write $u = v +x$ for some $x \in V$, $x \neq 0$. Thus, $$T(u) = T(u+x)$$
$$T(u) = T(u) + T(x)$$
$$T(u) + (-T(u)) = T(u) + T(x) + (-T(u))$$
$$0_{\tilde{V}} = T(x) + 0_{\tilde{V}}$$
$$0_{\tilde{V}} = T(x)$$
Since $\ker(T) = \{0_V\}$, it follows that $x = 0_V$. However this is a contradiction which implies that $u = v$.
Hence our assumption is incorrect and $T$ must be injective.
To prove that $T$ cannot be injective if $\ker(T) \neq \{0_V\}$, we take $k \in V$ such that $T(k) = 0_{\tilde{V}}$ and $k \neq 0_V$, then repeat a similar procedure till we get a contradiction.
Is this proof correct? Is there anything I am missing?
Best Answer
Here is a simple approach, suppose $KerT ={0}$ and suppose $T(u) = T(v)$ then it implies that $T(u-v) = 0$. Since kernel consists only $0$ vector we must have $u=v$. So $T$ is injective.
Suppose $T$ is injective then consider $u \in KerT$ and we know that $T(u) = 0 = T(0)$, since $T$ is injective we have $u=0$ and consequently $KerT$ consists only $0$ vector.