Prove that if $A$ is an $m$ x $n$ matrix and rank $A=n$, then the linear transformation $x\mapsto{}Ax$ is one-to-one.
I've just had a really hard time understanding onto and one-to-one, so I'm really not sure how to do this proof. Can anyone help me out?
A linear transformation $T$ is one-to-one if and only if $N(T) = \{0\}$.
Pf. Say $A$ is an $m$ x $n$ matrix with $rank(A)=n$. By Rank-Nullity, $rank(T)+nullity(T)=n$. Therefore, $nullity(T)=0$. Hence $N(T)=\{0\}$ and $T$ is one-to-one.
One to one means that $\forall x,y \ Ax=Ay \Rightarrow x=y$. This is equivalent to $A(x-y)=0 \Rightarrow x-y=0$, which is equivalent as well to $Az=0 \Rightarrow z=0$ which reads as Null(A)={0}. Use now the rank-nullity theorem.
As noted $T$ is one to one iff $ker(T)=\{0\}$. As $rank(T)=n$ by the rank-nullity theorem $dim(Ker(T))=0$ and the result follows. However this is a theorem, and not a definition. Let $T:V\to W$ be a linear transformation. $T$ is defined to be $1-1$ iff $$\forall u,v\in V ,T(u)=T(v)\implies u=v$$
Every $m\times n$ matrix $X$ satisfies $$ \DeclareMathOperator{rank}{rank}\rank(X^\top X)=\rank(X) $$ A proof can be found in this question.
In our case this fact implies that $A^\top A$ is an $n\times n$ matrix with rank $n$. It follows that $A^\top A$ is invertible.
Now, let $A_{\mathsf{L}}=(A^\top A)^{-1}A^\top$. Then $$ A_{\mathsf{L}}A=(A^\top A)^{-1}A^\top A=I $$ Hence $A\vec x=A\vec y$ implies $A_{\mathsf{L}}A\vec x=A_{\mathsf{L}}A\vec y$ so $I\vec x=I\vec y$ and $\vec x=\vec y$.
This actually gives a quick proof that every short exact sequence of linear maps between finite-dimensional vector spaces splits.
