This is the definition which we need in the proof of our theorem:
This is the theorem which we need in the proof of our theorem:
There is the theorem $9.5$:
A linear operator $A$ on a finite-dimensional vector space $X$ is one-to-one if and only if the range of $A$ is all of $X$.
There is the proof:
Let {$x_1,...,x_n$} be a basis of $X$. The linearity of $A$ shows that its range $\mathfrak R(A)$ is the span of the set $Q = {Ax_1,...,Ax_n}$. We therefore infer from Theorem $9.3(a)$ that $\mathfrak(A) = X$ if and only if $Q$ is independent. We have to prove that this happens if and only if $A$ is one-to-one.
Suppose $A$ is one-to-one and $\sum_{}c_i Ax_i = 0 $. Then $A(\sum_{}c_i x_i) = 0 $, hence $\sum_{} c_i x_i = 0 $, hence $c_1 = ... = c_n = 0$, and we conclude that $Q$ is independent.
Conversely, suppose $Q$ is independent and $A(\sum_{}c_i x_i) = 0$. Then $\sum_{}c_iAx_i = 0$, hence $c_1 = ... = c_n = 0$, and we conclude: $Ax = 0$ only if $x$ $=$ $0$. If now $Ax = Ay$, then $A(x-y)$ $=$ $Ax - Ay = 0 $, so that $x - y$ $=$ $0$, and this says that $A$ is one-to-one.
In the last part of theorem, I don't understand how do we conclude that $Ax = 0$ only if $x = 0$.
Any help would be appreciated.
$Ax = A(\sum_{}c_i x_i)... \implies c_i = 0 \implies x = \sum_{}c_i x_i = 0$