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SectionHSEHomogeneous Systems of Equations

In this section we specialize to systems of linear equations where every equation has a zero as its constant term. Along the way, we will begin to express more and more ideas in the language of matrices and begin a move away from writing out whole systems of equations. The ideas initiated in this section will carry through the remainder of the course.

SubsectionSHSSolutions of Homogeneous Systems

As usual, we begin with a definition.

DefinitionHSHomogeneous System

A system of linear equations, \(\linearsystem{A}{\vect{b}}\) is homogeneous if the vector of constants is the zero vector, in other words, if \(\vect{b}=\zerovector\text{.}\)

As you might have discovered by studying Example AHSAC, setting each variable to zero will always be a solution of a homogeneous system. This is the substance of the following theorem.

Proof

Since this solution is so obvious, we now define it as the trivial solution.

DefinitionTSHSETrivial Solution to Homogeneous Systems of Equations

Suppose a homogeneous system of linear equations has \(n\) variables. The solution \(x_1=0\text{,}\) \(x_2=0\text{,}\) …, \(x_n=0\) (i.e. \(\vect{x}=\zerovector\)) is called the trivial solution.

Here are three typical examples, which we will reference throughout this section. Work through the row operations as we bring each to reduced row-echelon form. Also notice what is similar in each example, and what differs.

After working through these examples, you might perform the same computations for the slightly larger example, Archetype J.

Notice that when we do row operations on the augmented matrix of a homogeneous system of linear equations the last column of the matrix is all zeros. Any one of the three allowable row operations will convert zeros to zeros and thus, the final column of the matrix in reduced row-echelon form will also be all zeros. So in this case, we may be as likely to reference only the coefficient matrix and presume that we remember that the final column begins with all zeros, and after any number of row operations is still all zeros.

Example HISAD suggests the following theorem.

Proof

Example HUSAB and Example HISAA are concerned with homogeneous systems where \(n=m\) and expose a fundamental distinction between the two examples. One has a unique solution, while the other has infinitely many. These are exactly the only two possibilities for a homogeneous system and illustrate that each is possible (unlike the case when \(n\gt m\) where Theorem HMVEI tells us that there is only one possibility for a homogeneous system).

SubsectionNSMNull Space of a Matrix

The set of solutions to a homogeneous system (which by Theorem HSC is never empty) is of enough interest to warrant its own name. However, we define it as a property of the coefficient matrix, not as a property of some system of equations.

DefinitionNSMNull Space of a Matrix

The null space of a matrix \(A\text{,}\) denoted \(\nsp{A}\text{,}\) is the set of all the vectors that are solutions to the homogeneous system \(\homosystem{A}\text{.}\)

In the Archetypes (Appendix A) each example that is a system of equations also has a corresponding homogeneous system of equations listed, and several sample solutions are given. These solutions will be elements of the null space of the coefficient matrix. We will look at one example.

Here are two (prototypical) examples of the computation of the null space of a matrix.

SubsectionReading Questions

1

What is always true of the solution set for a homogeneous system of equations?

2

Suppose a homogeneous system of equations has 13 variables and 8 equations. How many solutions will it have? Why?

3

Describe, using only words, the null space of a matrix. (So in particular, do not use any symbols.)

SubsectionExercises

C10

Each Archetype (Appendix A) that is a system of equations has a corresponding homogeneous system with the same coefficient matrix. Compute the set of solutions for each. Notice that these solution sets are the null spaces of the coefficient matrices.

Archetype A, Archetype B, Archetype C, Archetype D/Archetype E, Archetype F, Archetype G/Archetype H, Archetype I, Archetype J

C20

Archetype K and Archetype L are simply \(5\times 5\) matrices (i.e. they are not systems of equations). Compute the null space of each matrix.

For Exercises C21-C23, solve the given homogeneous linear system. Compare your results to the results of the corresponding exercise in Section TSS.

C21

\begin{align*} x_1 + 4x_2 + 3x_3 - x_4 &= 0\\ x_1 - x_2 + x_3 + 2x_4 &= 0\\ 4x_1 + x_2 + 6x_3 + 5x_4 &= 0 \end{align*}

Solution
C22

\begin{align*} x_1 - 2x_2 + x_3 - x_4 &= 0\\ 2x_1 - 4x_2 + x_3 + x_4 &= 0\\ x_1 - 2x_2 - 2x_3 + 3x_4 &= 0 \end{align*}

Solution
C23

\begin{align*} x_1 - 2x_2 + x_3 - x_4 &= 0\\ x_1 + x_2 + x_3 - x_4 &= 0\\ x_1 \quad + x_3 - x_4 &= 0 \end{align*}

Solution

For Exercises C25-C27, solve the given homogeneous linear system. Compare your results to the results of the corresponding exercise in Section TSS.

C25

\begin{align*} x_1 + 2x_2 + 3x_3 &= 0\\ 2x_1 - x_2 + x_3 &= 0\\ 3x_1 + x_2 + x_3 &= 0\\ \quad x_2 + 2x_3 &= 0 \end{align*}

Solution
C26

\begin{align*} x_1 + 2x_2 + 3x_3 &= 0\\ 2x_1 - x_2 + x_3 &= 0\\ 3x_1 + x_2 + x_3 &= 0\\ \quad 5x_2 + 2x_3 &= 0 \end{align*}

Solution
C27

\begin{align*} x_1 + 2x_2 + 3x_3 &= 0\\ 2x_1 - x_2 + x_3 &= 0\\ x_1 - 8x_2 - 7x_3 &= 0\\ \quad x_2 + x_3 &= 0 \end{align*}

Solution
C30

Compute the null space of the matrix \(A\text{,}\) \(\nsp{A}\text{.}\) \begin{equation*} A= \begin{bmatrix} 2 & 4 & 1 & 3 & 8 \\ -1 & -2 & -1 & -1 & 1 \\ 2 & 4 & 0 & -3 & 4 \\ 2 & 4 & -1 & -7 & 4 \end{bmatrix} \end{equation*}

Solution
C31

Find the null space of the matrix \(B\text{,}\) \(\nsp{B}\text{.}\) \begin{align*} B&= \begin{bmatrix} -6 & 4 & -36 & 6 \\ 2 & -1 & 10 & -1 \\ -3 & 2 & -18 & 3 \end{bmatrix} \end{align*}

Solution
M45

Without doing any computations, and without examining any solutions, say as much as possible about the form of the solution set for corresponding homogeneous system of equations of each archetype that is a system of equations.

Archetype A, Archetype B, Archetype C, Archetype D/Archetype E, Archetype F, Archetype G/Archetype H, Archetype I, Archetype J

For Exercises M50–M52 say as much as possible about each system's solution set. Be sure to make it clear which theorems you are using to reach your conclusions.

M50

A homogeneous system of 8 equations in 8 variables.

Solution
M51

A homogeneous system of 8 equations in 9 variables.

Solution
M52

A homogeneous system of 8 equations in 7 variables.

Solution
T10

Prove or disprove: A system of linear equations is homogeneous if and only if the system has the zero vector as a solution.

Solution
T11

Suppose that two systems of linear equations are equivalent. Prove that if the first system is homogeneous, then the second system is homogeneous. Notice that this will allow us to conclude that two equivalent systems are either both homogeneous or both not homogeneous.

Solution
T20

Consider the homogeneous system of linear equations \(\homosystem{A}\text{,}\) and suppose that \(\vect{u}\) is one solution to the system of equations. Prove that \(\vect{v}\) is also a solution to \(\homosystem{A}\text{.}\) \begin{align*} \vect{u}&=\colvector{u_1\\u_2\\u_3\\\vdots\\u_n}& \vect{v}&=\colvector{4u_1\\4u_2\\4u_3\\\vdots\\4u_n} \end{align*}

Solution