Connection between linear independence, non-/trivial and x solutions

Solution 1:

A set of vectors $\{v_1,\dots,v_n\}$ is linearly independent when the system of equations on $c_1,\dots,c_n$ given by $$ c_1 v_1 + c_2 v_2 + \cdots + c_n v_n = 0 $$ has only the solution $c_1=c_2 = \cdots = c_n = 0$. This solution is referred to as the trivial relation between these vectors, since no matter what vectors we choose, making $c_1=c_2 = \cdots = c_n = 0$ will aways make the other side $0$. So, this solution doesn't "distinguish" between sets of vectors.

We often say that a set of vectors is linearly dependent iff there exists a non-trivial relation between them; that is, iff there is a choice of $c_1,\dots,c_n$ in the above equation besides $c_1=c_2 = \cdots = c_n = 0$.

Now, suppose $A$ is a matrix whose columns are the vectors $v_1,\dots,v_n$. Then that first equation can be rewritten as $$ A \pmatrix{c_1\\ \vdots \\ c_n} = \pmatrix{0\\ \vdots \\ 0} $$ Or, in other words, $$ A \vec c = \vec 0 $$ This is sometimes called the homogeneous problem. This kind of problem will always have the trivial solution, $\vec c = \vec 0$. So, we can also say that the columns of a matrix are linearly independent if the associated homogeneous equation has only the trivial solution.

Note that in each of these equations, we have had a $0$ on the right side of the equation. Keep in mind that the matrix equation $$ Ax = \vec 0 $$ will always have at least one solution, since we always have the trivial solution $x = \vec 0$. If $\vec b$ is not necessarily the zero vector, then the equation $$ Ax = \vec b $$ can have either no solutions, one solution, or infinitely many. As it ends up, there is a useful way to write the solutions to this problem in terms of the solutions to the homogeneous problem.

Suppose that $x_0$ is a vector such that $A x_0 = \vec b$. Then, the every solution to the problem $A x = b$ can all be written in the form $x = x_0 + x_h$ for some vector $x_h$ such that $A x_h = 0$.

If any problem $A x = b$ has one solution, then the columns of $A$ must be linearly independent. If any such problem has infinitely many solutions, then the columns of $A$ must be linearly dependent.

If such a problem has no solution, we don't know whether or not the columns are independent. We do know, however, that $b$ is not a linear combination of the columns.