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?

  • 1
    $\begingroup$ 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. $\endgroup$ Commented Oct 22, 2014 at 11:58

2 Answers 2


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.

  • $\begingroup$ Is this imply that b is always linear combination of A , if there is a sol and infinitely many solutions? $\endgroup$
    – fiksx
    Commented Feb 28, 2018 at 3:46
  • $\begingroup$ @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. $\endgroup$ Commented Feb 28, 2018 at 16:27

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.

  1. 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; 
  1. 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;


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

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


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