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Homogeneous system

by , PhD

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


A homogenous system has the form[eq1]where A is a $K	imes L$ matrix of coefficients, x is a $L	imes 1$ vector of unknowns and 0 is the Kx1 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 system[eq4]where the matrix $R$ is in row echelon form (REF).

Trivial solution

A homogeneous system always has the solution[eq5]which is called trivial solution.

Basic and non-basic variables

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

Example Consider the following $2	imes 4$ 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 $K	imes L$ REF matrix $R$ has $r$ basic columns.

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

Partition the matrix $R$ into two blocks:[eq7]where $B$ is the $K	imes r$ sub-matrix of basic columns and $N$ is the [eq8] sub-matrix of non-basic columns.

Similarly, partition the vector of unknowns into two blocks:[eq9]where $x_{B}$ is the $r	imes 1$ vector of basic variables and $x_{N}$ is the [eq10] vector of non-basic variables.

Then, we can write the system of equations as[eq11]or[eq12]

General solution

The general solution of the homogeneous system[eq1]is the set of all possible solutions, that is, the set of all x 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 $B$ is full-rank and $rleq K$, the matrix [eq14] is full-rank (see the lecture on the rank of matrix products). Therefore, we can pre-multiply equation (1) by [eq15] so as to obtain[eq16]


Then, we have[eq18]

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 H whose coefficients are the non-basic variables:[eq20]

Thus, each column of H is a particular solution of the system, obtained by setting its corresponding non-basic variable equal to 1 and all the other non-basic variables equal to 0. 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 system[eq4]where[eq22]and[eq23]Then, we can define[eq24]The system can be written as[eq25]but since $B$ is the identity matrix, we have[eq26]Thus, the general solution of the system is the set of all vectors x that satisfy[eq27]

Solved exercises

Below you can find some exercises with explained solutions.

Exercise 1

Define the $3	imes 2$ matrix[eq28]

Find the general solution of the system[eq1]where x is a $2	imes 1$ vector of unknowns.


The matrix A 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 system[eq4]Since both of the two columns of $R$ are basic, there are no unknowns to choose arbitrarily. As a consequence, the only solution of the system is the trivial one ($x=0$).

Exercise 2

Define the $3	imes 2$ matrix[eq32]

Find the general solution of the system[eq1]where x is a $3	imes 1$ vector of unknowns.


For convenience, we are going to transform A into a reduced row echelon form matrix. We divide the second row by $2$; 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 system[eq4]The system can be written as[eq36]Thus, the general solution of the system is the set of all vectors x that satisfy[eq37]

How to cite

Please cite as:

Taboga, Marco (2021). "Homogeneous system", Lectures on matrix algebra.

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