Is this a valid technique for this linear algebra proof? Let {$A$v$_1$....$A$v$_n$} be a linearly independent set. If $A$ is invertible, prove that {v$_1$....v$_n$} is linearly independent.
PROOF:
Because {$A$v$_1$....$A$v$_n$} is a linearly independent set,
the equation $c_1A$v$_1$+....+$c_nA$v$_n=0$ implies that $c_1$....$c_n$ must all equal $0$.
multiplying the equation by $A^{-1}$ we obtain: $c_1$v$_1$+....+$c_n$v$_n=0$, where $c_1$....$c_n$ must still must all be $0$.
We can therefore conclude that {v$_1$....v$_n$} is linearly independent
 A: I see a bit of a problem with the argument. To prove that a set is linearly independent, you have to allow $c_1...c_n$ to be an arbitrary tuple that yields the zero vector when used as coefficients for the set, then prove they're all zero. That is, you need $c_1 \textbf v_1 +...+c_n\textbf v_n =\textbf 0 \rightarrow c_1...c_n = 0$. You have instead proved $c_1 \textbf{Av}_1 +...+c_n\textbf{Av}_n =\textbf 0 \rightarrow c_1...c_n = 0$. Now, it certainly is true that if $c_1 \textbf{Av}_1 +...+c_n\textbf{Av}_n = \textbf 0$, then $c_1 \textbf v_1 +...+c_n\textbf v_n =\textbf 0$, so if you have all $c_1...c_n$ for   which $c_1 \textbf{Av}_1 +...+c_n\textbf{Av}_n = \textbf 0$, then all of your tuples are tuples that you are looking for. But "everything I find is a tuple I'm looking for" and "all I'm looking for is a tuple I find" are different statement.   If there is any set of coefficients for which $c_1 \textbf v_1 +...+c_n\textbf v_n =\textbf 0$ but $c_1 \textbf{Av}_1 +...+c_n\textbf{Av}_n \neq \textbf 0$, then you haven't proven that $c_1...c_n=0$ for all $c_1...c_n$ for which $c_1 \textbf v_1 +...+c_n\textbf v_n =\textbf 0$. There aren't actually any such coefficients, and this fact is relatively trivial, but it's about the same level of triviality as what you're being asked to prove, so it isn't really valid to assume it.
When doing proofs, it's important to follow quantifiers exactly.
