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?
 A: 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. 
A: It seems c is constant in Ben's context. So the exact expression of(c1,...,cn)^T should be changed to (v1,...vn)^T. Please see the following changes.
Original form in Ben's context:
Ac^T = (0,...,0)^T 

is replaced by the right expression as follows.
Av^T = (0,...,0)^T

In addition, b is not vector but scalar. So the mathematical arrow symbol of vector should be deleted in order to address the right context.

*

*Homogeneous Linear System(HLS) - Ternary Form

With regard to the following homogenous linear system Ax = 0, A is a coefficient matrix,
a11x1 + a12x2 + a13x3 = 0,
a21x1 + a22x2 + a23x3 = 0,
a31x1 + a32x2 + a33x3 = 0;

The homogeneous linear system has the two kinds of solutions.
1). zero solution
The sufficient and necessary condition of the zero solution for is listed as follows:
det(A) ≠ 0;

2). Non-zero solution:
The sufficient and necessary condition of the zero solution is listed as follow:
det(A) = 0; 



*Non-Homogenous(NHLS) - Ternary Form:

With regard to the non-homogenous linear system Ax = b while A is an augmented matrix including b (in contrast with the above-mentioned coefficient matrix), it has the standard ternary linear form form.
a11x1 + a12x2 + a13x3 = b1,
a21x1 + a22x2 + a23x3 = b2,
a31x1 + a32x2 + a33x3 = b3;

Please remember aij is co-efficient(that forms the co-efficient matrix A), xi(or x1,x2,x3) is vector and bi(or b1,b2,b3) is scalar.
We will get the three scenarios as follows.
1). Unique solution:
We will get a unique solution after reducing the above-mentioned matrix A(that is an augmented form) to the strictly upper triangular matrix U that is the echelon form with all zero items following after the non-zero diagonal items according to the elimination method of Gauss-Jordan (or LU decomposition).
2). Infinitely many solutions
We will get infinitely many solutions after reducing the (augmented) matrix A to the following form.
either the combination as follows:
x1 + c13x3 = d1,
x2 + c23x3 = d2,
0 = 0;

where both d1 and d2 are scalars after using element-elimination operation(s).
or the following form as follows:
x1 + c12x2 + c13x3 = d1,
0 = 0,
0 = 0;

3). No solution
We will get no solution after reducing the original coefficient matrix(or non-augmented matrix) with b = 0 to the variant forms as follows;
either the reduced combination (one of them have one zero item):
x1 + c13x3 = d1,
x2 + c23x3 = d2,
0 = d3;

where d1, d2 and d3 are obtained scalars after using element-elimination operations.
or the reduced combinations (with two zeros):
x1 + c12x2 + c13x3 = d1,
0 = d2,
0 = 0;

Notes:

*

*The linear algebra is quite abstract, but it is the essence of the matrix arithmetic.


*Any first listed equation in the ternary equation systems should be a monic polynomial.
