Show that $S \in \operatorname{Hom}(\textsf V , \textsf V)$ is invertible if and only if whenever $v_1,\dots,v_n \in \textsf V$ are linearly independent, then $S(v_1),\dots,S(v_n)$ are also linearly independent.
Here $S$ is invertible means that the constant term of the minimal polynomial for $S$ is not $0$.
I need help to prove that when $S(v_1),\dots,S(v_n)$ is linearly independent and $v_1,\dots,v_n$ are also linearly independent then $S$ is invertible because I have proven that if $S$ is regular and $v_1,\dots,v_n$ are linearly independent then $S(v_1),\dots,S(v_n)$ are also linearly independent.
Thanks in advance.
Best Answer
Suppose $\{v_i\}$ is a basis for $V$, then so is $\{S(v_i)\}$, by assumption.
Let $v = \sum a_i v_i \in V$ be an arbitrary vector, and write $v = \sum c_i S(v_i) = \sum S(c_iv_i)=S(\sum c_iv_i)$ by linearity. So, $S$ is onto.
Suppose $S(v)=S(w)$. Then write $v=\sum a_iv_i$ and $w=\sum b_iv_i$. We have: $S(\sum a_iv_i) = S(\sum b_iv_i)$ or
$$\sum (a_i-b_i)S(v_i) = 0$$
Since $S(v_i)$'s are linearly independent, $a_i=b_i$ for each $i$, i.e. $v=w$ so that $S$ is injective.