My book defines an isometry as a linear operator between two vector spaces X and Y where:
$$\|T(x)\|=\|x\|$$
Later it has a sentence which I do not understand.
If we have a finite dimensional normed vector space where $T$ is an isometry of $X$ into $X$, then $T$ is also onto.
How do we know that it is also onto?
A direct proof, in response to your comment:
Let $T$ be an isometry, and let $v_1,\dots,v_n$ be a basis of $X$. We note that the vectors $T(v_1),\dots,T(v_n)$ must be linearly independent.
Why? Suppose otherwise. That is, suppose that $$ \sum_{i=1}^n a_i T(v_i) = 0 $$ for some choice of $a_1,\dots,a_n$ not all zero. It follows that $$ T\left(\sum_{i=1}^n a_i v_i\right) = 0 $$ So, we have $\left\|\sum_{i=1}^n a_i v_i\right\| \neq 0$ but $\|T\left(\sum_{i=1}^n a_i v_i\right)\| = 0$, which is a contradiction. (It is also possible to get a contradiction from the injectivity of $T$).
So, $T(v_1),\dots,T(v_n)$ is a set of $n$ linearly independent vectors in $X$, which means that these vectors form a basis of $X$.
So, let $v \in X$ be arbitrarily chosen. There exist $a_i$ so that $v = \sum_{i=1}^n a_iT(v_i)$. It follows that $v = T(\sum_{i=1}^n a_i v_i)$.
So, $T$ is indeed onto.