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

I am having a hard time remembering which goes hand in hand with what. The math questions I get always include words like trivial etc.

• 1 solution
• no solution
• infinite amount of solutions

And then we have the two types of set of vectors

• linearly independent
• linearly dependent

A set of vectors is linearly independent iff the system of equations are satisfied when all vector scalars are = 0 (making all vectors zero vectors). This results in 1 solution, the solution is trivial?

A set of vectors is linearly dependent when there are an infinite amount of solutions to the system of equations. This is non-trivial?

Where does no solution come in? I understand that if there is no solution, then all of the vectors do not intersect at a specific coordinate(which is the solution to the system of equations). But does that mean the set of vectors is linearly independent or dependent?

• It is not typical to say that "a matrix is linearly dependent/independent". Normally, we say that a set of vectors or the rows/columns of a matrix are linearly independent/dependent. – Omnomnomnom Oct 22 '14 at 11:58

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.

• Is this imply that b is always linear combination of A , if there is a sol and infinitely many solutions? – fiksx Feb 28 '18 at 3:46
• @Vixf if $Ax = b$ has at least one solution, then $b$ is a linear combination of the columns of $A$. I'm not quite sure I understood your question, but I hope that answers it. – Omnomnomnom Feb 28 '18 at 16:27