Homogeneous system

A homogeneous system of equations is a system in which the vector of constants on the right-hand side of the equals sign is zero.

In this lecture we provide a general characterization of the set of solutions of a homogeneous system.

Table of contents

Definition
Equivalent system in row echelon form
Trivial solution
Basic and non-basic variables
Partitioned system
General solution
Solved exercises
1. Exercise 1
2. Exercise 2

Definition

A homogenous system has the formwhere is a matrix of coefficients, is a vector of unknowns and is the zero vector.

Example The system [eq2] which can be written in matrix form as [eq3] is homogeneous.

Equivalent system in row echelon form

By performing elementary row operations on a homogenous system, we obtain equivalent systems that are all homogenous. In fact, elementary row operations (multiplying an equation by a non-zero constant; adding a multiple of one equation to another equation; interchanging two equations) leave the zero vector of constants on the right-hand side of the equals sign unaffected.

As a consequence, we can transform the original system into an equivalent homogeneous systemwhere the matrix is in row echelon form (REF).

Trivial solution

A homogeneous system always has the solutionwhich is called trivial solution.

Basic and non-basic variables

Remember that the columns of a REF matrix are of two kinds:

basic columns: they contain a pivot (i.e., a non-zero entry such that we find only zero entries in the quadrant starting from the pivot and extending below it and to its left);
non-basic columns: they do not contain a pivot.

Example Consider the following matrix in row echelon form: [eq6] The first and the third columns are basic, while the second and the fourth are non-basic.

Partitioned system

Suppose that the REF matrix has basic columns.

Without loss of generality, we can assume that the first columns are basic and the last are non-basic (we can re-number the unknowns if necessary).

Partition the matrix into two blocks:where is the sub-matrix of basic columns and is the sub-matrix of non-basic columns.

Similarly, partition the vector of unknowns into two blocks: [eq9] where $x_{B}$ is the vector of basic variables and $x_{N}$ is the vector of non-basic variables.

Then, we can write the system of equations as [eq11] or

General solution

The general solution of the homogeneous systemis the set of all possible solutions, that is, the set of all that satisfy the system of equations.

We already know that, if the system has a solution, then we can arbitrarily choose the values of the non-basic variables $x_{N}$ and then find, by the back-substitution algorithm, the values of the basic variables $x_{B}$ that solve the system. In the homogeneous case, the existence of a solution is not an issue because the vector of constants is zero (revise the lecture on the row echelon form if you are wondering why).

Therefore, there is a unique $x_{B}$ that solves equation (1) for any arbitrary choice of $x_{N}$ .

Since is full-rank and , the matrix is full-rank (see the lecture on the rank of matrix products). Therefore, we can pre-multiply equation (1) by so as to obtain

Define

Then, we have

The latter can be used to characterize the general solution of the homogeneous system: it explicitly links the values of the basic variables to those of the non-basic variables that can be set arbitrarily.

Denote by $S_{H}$ the general solution (i.e., the set of all possible solutions). Then, we have [eq19]

The product $Hx_{N}$ can be seen as a linear combination of the columns of whose coefficients are the non-basic variables:

Thus, each column of is a particular solution of the system, obtained by setting its corresponding non-basic variable equal to and all the other non-basic variables equal to . By taking linear combination of these particular solutions, we obtain the general solution.

Clearly, the general solution embeds also the trivial one, which is obtained by setting all the non-basic variables to zero.

Example Consider the homogeneous systemwhere [eq22] and [eq23] Then, we can define [eq24] The system can be written asbut since is the identity matrix, we haveThus, the general solution of the system is the set of all vectors that satisfy [eq27]

Solved exercises

Below you can find some exercises with explained solutions.

Exercise 1

Define the matrix [eq28]

Find the general solution of the systemwhere is a vector of unknowns.

Solution

The matrix is not in row echelon form, but we can subtract three times the first row from the third one in order to obtain an equivalent matrix in row echelon form: [eq30] Thus, we can discuss the solutions of the equivalent systemSince both of the two columns of are basic, there are no unknowns to choose arbitrarily. As a consequence, the only solution of the system is the trivial one ().

Exercise 2

Define the matrix [eq32]

Find the general solution of the systemwhere is a vector of unknowns.

Solution

For convenience, we are going to transform into a reduced row echelon form matrix. We divide the second row by ; then, we subtract two times the second row from the first one. The result is an equivalent matrix in reduced row echelon form: [eq34] We can now discuss the solutions of the equivalent systemThe system can be written as [eq36] Thus, the general solution of the system is the set of all vectors that satisfy [eq37]

How to cite

Please cite as:

Taboga, Marco (2021). "Homogeneous system", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/homogeneous-system.