Contrapositive arguments I came across the following problem:-

Let $S$= {${u_{1},u_{2},...u_{n}}$} $\subseteq  \Bbb C^{n}\  $and $T$= {${Au_{1},Au_{2},...Au_{n}}$}, for some matrix square matrix A $\in$$\Bbb M_{n}($C$)$.
If $S$ is Linearly independent prove that T is linearly independent for every invertible matrix A.

I tried to prove it using contrapositive argument. But I do not understand how to do it. I have two arguments running parallel.
(i) Let T is linearly dependent then S is linearly dependent for some invertible matrix A
(ii) Let T is linearly dependent then S is linearly dependent for some non-invertible matrix A
Which of the following is the correct one?
It would be a great help if someone provides some insight as to how do we, in general, do contrapositive of an argument like this.
 A: First of all, I find that this statement is easier to prove directly rather than by contrapositive. In particular, if $S$ is linearly independent, then we use the definition of linear dependence to show that $T$ is also linearly independent. If $c_1,\dots, c_n$ are such that $c_1 A v_1 + \cdots + c_n A v_n = 0$, then
$$
c_1 A v_1 + \cdots + c_n A v_n = 0 \implies\\
A(c_1 v_1 + \cdots + c_n v_n) = 0 \implies\\
c_1 v_1 + \cdots + c_n v_n \implies\\
c_1 = \cdots = c_n = 0,
$$
as desired.
As for the contrapositive of the statement to be proved: (i) is the correct version of the contrapositive.
Here is a justification. Let $p$ denote the statement "$S$ is linearly independent", and let $q$ denote the statement "$T_A$ is linearly independent for every invertible matrix $A$"; I have used the notation $T_A$ to emphasize that the set $T$ depends on our choice of $A$. The statement that you are trying to prove is $p \implies q$; the contrapositive of this would be $\lnot q \implies \lnot p$.  It is clear that $\lnot p$ is simlply "$S$ is linearly dependent", but deciding what $\lnot q$ should be is trickier.
We can break the statement $q$ down into "for every invertible matrix $A$: $T_A$ is linearly indpendent". If we let $r_A$ denote the statement "$T_A$ is linearly independent", then we can write this symbolically as
$$
q = \forall A \in GL_n:r_A.
$$
Here, $A \in GL_n$ means "$A$ is invertible". Now, if it is not true that "for all $A \in GL_n$, $r_A$ is true", then there must be a counterexample. That is, there must be an example of an $A \in GL_n$ for which $r_A$ is not true. Symbolically, we might write
$$
\lnot (\forall A \in GL_n:r_A) = \exists A \in GL_n : \lnot r_A.
$$

Analogously, if we wanted to disprove the statement "every swan is white" (i.e. "for every swan, the color of the swan is white"), then we would have to find a swan that is not white. So, the negation is "there exists a non-white swan" (i.e. "there exists a swan such that the color of the swan is not white).
In this context, your statement (ii) is the equivalent of saying "there exists a non-swan that is not white."
