I have difficulty proving this:
Let $T:\mathbb R^n \to\mathbb R^n$ be a linear transformation with standard matrix $A$. Prove that $T$ is one-to-one if and only if $A$ is invertible.(Show both directions, but not using the Invertible Matrix Theorem)
I know that $T$ is one-to-one if and only if the columns of $A$ are linearly Independent.
So what I was thinking is $Ax=0$ since this will show that its linearly independent, but I do not know if I am in the right path, also I don't know what it means with both directions so I am stock.
Any help, thanks
Best Answer
"$\Rightarrow$" First, suppose that $T$ is one-to-one. We'll prove that A is invertible by showing that $Ax = 0$ has only the trivial solution $x=0$. It suffices to show that $\ker(A)$ has only one element. So suppose there exist $x, x'$ in $\ker(A)$ with $x \neq x'$. Since $T$ is one-to-one, then $Ax \neq Ax'$, but $x, x' \in \ker(A)$ implies $0 = Ax = Ax'$, a contradiction.
"$\Leftarrow$" Now suppose that $A$ is invertible and prove that $T$ is one-to-one. Let $x, 'x \in \mathbb{R}^n$ such that $T(x)=T(x')$, that is, $Ax = Ax'$. Then $Ax-Ax'=A(x-x')=0$, so that $x-x' \in \ker(A)$. As $A$ is invertible, $\ker(A)=\{0\}$. Thus $x-x'=0$ so we have $x=x'$ and $T$ is one-to-one.