If a matrix does not have have only the trivial solution, are the columns linearly dependent? I know :
"The columns of a matrix A are linearly independent if and only if the equation $$Ax=0$$ has only the trivial solution."
What if the equation has a nontrivial solution. Does this imply that the columns are linearly dependant? I have a hard time with if and only if statements. 
 A: Yes exactly, this is logic. If $p$ and $q$ are two propositions and $p$ implies $q$ is true, then the negation of $q$ implies the negation of $p$.
A: Yes. Whenever you have an "if and only if" statement, four implications follow from it:


*

*The forward direction, $P \implies Q$. "If the columns of $A$ are linearly independent, then the equation $Ax = 0$ has only the trivial solution."

*The contrapositive of the forward direction, $\neg Q \implies \neg P$. "If the equation $Ax = 0$ has a nontrivial solution, then the columns of $A$ are linearly dependent."

*The backward direction, $Q \implies P$. "If the equation $Ax=0$ has only the trivial solution, then the columns of $A$ are linearly independent."

*The contrapositive of the backward direction, $\neg P \implies \neg Q$. "If the columns of $A$ are linearly dependent, then the equation $Ax=0$ has a nontrivial solution."


Here, 1 and 2 are logically equivalent, as are 3 and 4. So if you were proving the if and only if statement, proving either one of 1 and 2, and then proving either one of 3 and 4, would be fine. If you're using the statement, you get to use whichever of the implications 1, 2, 3, 4 is relevant.
A: HINT
Simply note that
$$A\vec x=\sum x_i\vec v_i=0$$
represents a linear  combination of the column vectors of $A$ by the coefficient of $$\vec x=(x_1,x_2,...,x_n)$$
then recall the definition of linearly independent/dependent vectors.
