prove for T:V→V is linear, if surjective then T is bijective

Question

I know that $T$ is surjective means $R(T)=V$, which means $\mathrm{Nullity}(T)=0$.

But how can I show that $\mathrm{Nullity}(T)=0$ implies $T$ is injective?

I know that if $f(x)=f(y)$ and $x=y$ then $T$ is one-to-one. Do I show $T(0)=0$, $0$ is the only vector in $V$ that can let $T$ maps it into $0$? But my question is that even if I showed above, how can we be sure that $T(0)=0$ indicates that for any other vectors the linear map is one-to-one ?

I feel like that I got caught in a loop or missed something here. Can anyone let me know what part I'm not considering in? Thanks.

Answer 1

You say you know that Nullity$(T)=\{0\}$. Suppose that $T(x) = T(y)$. Then $T(x-y) = 0$, so $x-y=0$ which implies $x=y$.

In fact, this is a common technique with linear transformations; to show that $T$ is injective you show that the Nullity is $\{0 \}$.

Answer 2

Here's a hint. Start by assuming that $T(x) = T(y)$ for some $x$ and $y$. Subtracting both sides by $T(y)$, we have $T(x) - T(y) = 0$. Use linearity to get $T(x-y) = 0$. Now use the fact that $0$ is the only vector that get mapped to $0$. Hope this helps!