Section HSE Homogeneous Systems of Equations

Subsection SHS: Solutions of Homogeneous Systems

As usual, we begin with a definition.

Definition HS
Homogeneous System
A system of linear equations, $ℒ S (A, b)$ is homogeneous if the vector of constants is the zero vector, in other words, $b = 0$ . $△$

Example AHSAC
Archetype C as a homogeneous system
For each archetype that is a system of equations, we have formulated a similar, yet different, homogeneous system of equations by replacing each equation’s constant term with a zero. To wit, for Archetype C, we can convert the original system of equations into the homogeneous system,

$\begin{array}{l} 2 x_{1} - 3 x_{2} + x_{3} - 6 x_{4} & = 0 \\ 4 x_{1} + x_{2} + 2 x_{3} + 9 x_{4} & = 0 \\ 3 x_{1} + x_{2} + x_{3} + 8 x_{4} & = 0 \end{array}$

Can you quickly find a solution to this system without row-reducing the augmented matrix? $⊠$

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.

Theorem HSC
Homogeneous Systems are Consistent
Suppose that a system of linear equations is homogeneous. Then the system is consistent. $□$

Proof Set each variable of the system to zero. When substituting these values into each equation, the left-hand side evaluates to zero, no matter what the coefficients are. Since a homogeneous system has zero on the right-hand side of each equation as the constant term, each equation is true. With one demonstrated solution, we can call the system consistent. $■$

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

Definition TSHSE
Trivial Solution to Homogeneous Systems of Equations
Suppose a homogeneous system of linear equations has $n$ variables. The solution $x_{1} = 0$ , $x_{2} = 0$ ,…, $x_{n} = 0$ (i.e. $x = 0$ ) 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.

Example HUSAB
Homogeneous, unique solution, Archetype B
Archetype B can be converted to the homogeneous system,

$\begin{array}{l} - 11 x_{1} + 2 x_{2} - 14 x_{3} & = 0 \\ 23 x_{1} - 6 x_{2} + 33 x_{3} & = 0 \\ 14 x_{1} - 2 x_{2} + 17 x_{3} & = 0 \end{array}$

whose augmented matrix row-reduces to

$[\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{matrix}]$

By Theorem HSC, the system is consistent, and so the computation $n - r = 3 - 3 = 0$ means the solution set contains just a single solution. Then, this lone solution must be the trivial solution. $⊠$

Example HISAA
Homogeneous, infinite solutions, Archetype A
Archetype A can be converted to the homogeneous system,

$\begin{array}{l} x_{1} - x_{2} + 2 x_{3} & = 0 \\ 2 x_{1} + x_{2} + x_{3} & = 0 \\ x_{1} + x_{2} & = 0 \end{array}$

whose augmented matrix row-reduces to

$[\begin{matrix} 1 & 0 & 1 & 0 \\ 0 & 1 & - 1 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}]$

By Theorem HSC, the system is consistent, and so the computation $n - r = 3 - 2 = 1$ means the solution set contains one free variable by Theorem FVCS, and hence has infinitely many solutions. We can describe this solution set using the free variable $x_{3}$ ,

$S = \{[\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}]| x_{1} = - x_{3}, x_{2} = x_{3}\} = \{[\begin{matrix} - x_{3} \\ x_{3} \\ x_{3} \end{matrix}]| x_{3} \in ℂ^{}\}$

Geometrically, these are points in three dimensions that lie on a line through the origin. $⊠$

Example HISAD
Homogeneous, infinite solutions, Archetype D
Archetype D (and identically, Archetype E) can be converted to the homogeneous system,

$\begin{array}{l} 2 x_{1} + x_{2} + 7 x_{3} - 7 x_{4} & = 0 \\ - 3 x_{1} + 4 x_{2} - 5 x_{3} - 6 x_{4} & = 0 \\ x_{1} + x_{2} + 4 x_{3} - 5 x_{4} & = 0 \end{array}$

whose augmented matrix row-reduces to

$[\begin{matrix} 1 & 0 & 3 & - 2 & 0 \\ 0 & 1 & 1 & - 3 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}]$

By Theorem HSC, the system is consistent, and so the computation $n - r = 4 - 2 = 2$ means the solution set contains two free variables by Theorem FVCS, and hence has infinitely many solutions. We can describe this solution set using the free variables $x_{3}$ and $x_{4}$ ,

$\begin{array}{l} S & = \{[\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{matrix}]| x_{1} = - 3 x_{3} + 2 x_{4}, x_{2} = - x_{3} + 3 x_{4}\} \\ = \{[\begin{matrix} - 3 x_{3} + 2 x_{4} \\ - x_{3} + 3 x_{4} \\ x_{3} \\ x_{4} \end{matrix}]| x_{3}, x_{4} \in ℂ^{}\} \end{array}$

$⊠$

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 zeros, and after any number of row operations is still zero.

Example HISAD suggests the following theorem.

Theorem HMVEI
Homogeneous, More Variables than Equations, Infinite solutions
Suppose that a homogeneous system of linear equations has $m$ equations and $n$ variables with $n > m$ . Then the system has infinitely many solutions. $□$

Proof We are assuming the system is homogeneous, so Theorem HSC says it is consistent. Then the hypothesis that $n > m$ , together with Theorem CMVEI, gives infinitely many solutions. $■$

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 $n > m$ where Theorem HMVEI tells us that there is only one possibility for a homogeneous system).

Subsection NSM: Null 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.

Definition NSM
Null Space of a Matrix
The null space of a matrix $A$ , denoted $N (A)$ , is the set of all the vectors that are solutions to the homogeneous system $ℒ S (A, 0)$ .

(This definition contains Notation NSM.) $△$

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’ll look at one example.

Example NSEAI
Null space elements of Archetype I
The write-up for Archetype I lists several solutions of the corresponding homogeneous system. Here are two, written as solution vectors. We can say that they are in the null space of the coefficient matrix for the system of equations in Archetype I.

$\begin{array}{l} x = [\begin{matrix} 3 \\ 0 \\ - 5 \\ - 6 \\ 0 \\ 0 \\ 1 \end{matrix}] & y = [\begin{matrix} - 4 \\ 1 \\ - 3 \\ - 2 \\ 1 \\ 1 \\ 1 \end{matrix}] \end{array}$

However, the vector

$z = [\begin{matrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 2 \end{matrix}]$

is not in the null space, since it is not a solution to the homogeneous system. For example, it fails to even make the first equation true. $⊠$

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

Example CNS1
Computing a null space, #1
Let’s compute the null space of

$A = [\begin{matrix} 2 & - 1 & 7 & - 3 & - 8 \\ 1 & 0 & 2 & 4 & 9 \\ 2 & 2 & - 2 & - 1 & 8 \end{matrix}]$

which we write as $N (A)$ . Translating Definition NSM, we simply desire to solve the homogeneous system $ℒ S (A, 0)$ . So we row-reduce the augmented matrix to obtain

$[\begin{matrix} 1 & 0 & 2 & 0 & 1 & 0 \\ 0 & 1 & - 3 & 0 & 4 & 0 \\ 0 & 0 & 0 & 1 & 2 & 0 \end{matrix}]$

The variables (of the homogeneous system) $x_{3}$ and $x_{5}$ are free (since columns 1, 2 and 4 are pivot columns), so we arrange the equations represented by the matrix in reduced row-echelon form to

$\begin{array}{l} x_{1} & = - 2 x_{3} - x_{5} \\ x_{2} & = 3 x_{3} - 4 x_{5} \\ x_{4} & = - 2 x_{5} \end{array}$

So we can write the infinite solution set as sets using column vectors,

$N (A) = \{[\begin{matrix} - 2 x_{3} - x_{5} \\ 3 x_{3} - 4 x_{5} \\ x_{3} \\ - 2 x_{5} \\ x_{5} \end{matrix}]| x_{3}, x_{5} \in ℂ^{}\}$

$⊠$

Example CNS2
Computing a null space, #2
Let’s compute the null space of

$C = [\begin{matrix} - 4 & 6 & 1 \\ - 1 & 4 & 1 \\ 5 & 6 & 7 \\ 4 & 7 & 1 \end{matrix}]$

which we write as $N (C)$ . Translating Definition NSM, we simply desire to solve the homogeneous system $ℒ S (C, 0)$ . So we row-reduce the augmented matrix to obtain

$[\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}]$

There are no free variables in the homogeneous system represented by the row-reduced matrix, so there is only the trivial solution, the zero vector, $0$ . So we can write the (trivial) solution set as

$N (C) = \{0\} = \{[\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]\}$

$⊠$

Subsection READ: Reading Questions

Subsection EXC: Exercises

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
and Archetype J
Contributed by Robert Beezer

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.
Contributed by Robert Beezer

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

$\begin{array}{l} x_{1} + 4 x_{2} + 3 x_{3} - x_{4} & = 0 \\ x_{1} - x_{2} + x_{3} + 2 x_{4} & = 0 \\ 4 x_{1} + x_{2} + 6 x_{3} + 5 x_{4} & = 0 \end{array}$

Contributed by Chris Black Solution [ 208]

C22

$\begin{array}{l} x_{1} - 2 x_{2} + x_{3} - x_{4} & = 0 \\ 2 x_{1} - 4 x_{2} + x_{3} + x_{4} & = 0 \\ x_{1} - 2 x_{2} - 2 x_{3} + 3 x_{4} & = 0 \end{array}$

Contributed by Chris Black Solution [ 209]

C23

$\begin{array}{l} x_{1} - 2 x_{2} + x_{3} - x_{4} & = 0 \\ x_{1} + x_{2} + x_{3} - x_{4} & = 0 \\ x_{1} + x_{3} - x_{4} & = 0 \end{array}$

Contributed by Chris Black Solution [ 210]

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

$\begin{array}{l} x_{1} + 2 x_{2} + 3 x_{3} & = 0 \\ 2 x_{1} - x_{2} + x_{3} & = 0 \\ 3 x_{1} + x_{2} + x_{3} & = 0 \\ x_{2} + 2 x_{3} & = 0 \end{array}$

Contributed by Chris Black Solution [ 211]

C26

$\begin{array}{l} x_{1} + 2 x_{2} + 3 x_{3} & = 0 \\ 2 x_{1} - x_{2} + x_{3} & = 0 \\ 3 x_{1} + x_{2} + x_{3} & = 0 \\ 5 x_{2} + 2 x_{3} & = 0 \end{array}$

Contributed by Chris Black Solution [ 211]

C27

$\begin{array}{l} x_{1} + 2 x_{2} + 3 x_{3} & = 0 \\ 2 x_{1} - x_{2} + x_{3} & = 0 \\ x_{1} - 8 x_{2} - 7 x_{3} & = 0 \\ x_{2} + x_{3} & = 0 \end{array}$

Contributed by Chris Black Solution [ 212]

C30 Compute the null space of the matrix $A$ , $N (A)$ .

$A = [\begin{matrix} 2 & 4 & 1 & 3 & 8 \\ - 1 & - 2 & - 1 & - 1 & 1 \\ 2 & 4 & 0 & - 3 & 4 \\ 2 & 4 & - 1 & - 7 & 4 \end{matrix}]$

Contributed by Robert Beezer Solution [ 213]

C31 Find the null space of the matrix $B$ , $N (B)$ .

$\begin{array}{l} B & = [\begin{matrix} - 6 & 4 & - 36 & 6 \\ 2 & - 1 & 10 & - 1 \\ - 3 & 2 & - 18 & 3 \end{matrix}] \end{array}$

Contributed by Robert Beezer Solution [ 215]

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
Contributed by Robert Beezer

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.
Contributed by Robert Beezer Solution [ 216]

M51 A homogeneous system of 8 equations in 9 variables.
Contributed by Robert Beezer Solution [ 216]

M52 A homogeneous system of 8 equations in 7 variables.
Contributed by Robert Beezer Solution [ 216]

T10 Prove or disprove: A system of linear equations is homogeneous if and only if the system has the zero vector as a solution.
Contributed by Martin Jackson Solution [ 216]

T12 Give an alternate proof of Theorem HSC that uses Theorem RCLS.
Contributed by Ivan Kessler

T20 Consider the homogeneous system of linear equations $ℒ S (A, 0)$ , and suppose that $u = [\begin{matrix} u_{1} \\ u_{2} \\ u_{3} \\ ⋮ \\ u_{n} \end{matrix}]$ is one solution to the system of equations. Prove that $v = [\begin{matrix} 4 u_{1} \\ 4 u_{2} \\ 4 u_{3} \\ ⋮ \\ 4 u_{n} \end{matrix}]$ is also a solution to $ℒ S (A, 0)$ .
Contributed by Robert Beezer Solution [ 217]

Subsection SOL: Solutions

C21 Contributed by Chris Black Statement [ 201]
The augmented matrix for the given linear system and its row-reduced form are:

$\begin{array}{l} [\begin{matrix} 1 & 4 & 3 & - 1 & 0 \\ 1 & - 1 & 1 & 2 & 0 \\ 4 & 1 & 6 & 5 & 0 \end{matrix}] & \to_{}^{RREF} [\begin{matrix} 1 & 0 & 7 ∕ 5 & 7 ∕ 5 & 0 \\ 0 & 1 & 2 ∕ 5 & - 3 ∕ 5 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}] . \end{array}$

Thus, we see that the system is consistent (as predicted by Theorem HSC) and has an infinite number of solutions (as predicted by Theorem HMVEI). With suitable choices of $x_{3}$ and $x_{4}$ , each solution can be written as

$\begin{array}{l} [\begin{matrix} - \frac{7}{5} x_{3} - \frac{7}{5} x_{4} \\ - \frac{2}{5} x_{3} + \frac{3}{5} x_{4} \\ x_{3} \\ x_{4} \end{matrix}] \end{array}$

C22 Contributed by Chris Black Statement [ 202]
The augmented matrix for the given linear system and its row-reduced form are:

$\begin{array}{l} [\begin{matrix} 1 & - 2 & 1 & - 1 & 0 \\ 2 & - 4 & 1 & 1 & 0 \\ 1 & - 2 & - 2 & 3 & 0 \end{matrix}] & \to_{}^{RREF} [\begin{matrix} 1 & - 2 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{matrix}] . \end{array}$

Thus, we see that the system is consistent (as predicted by Theorem HSC) and has an infinite number of solutions (as predicted by Theorem HMVEI). With a suitable choice of $x_{2}$ , each solution can be written as

$\begin{array}{l} [\begin{matrix} 2 x_{2} \\ x_{2} \\ 0 \\ 0 \end{matrix}] \end{array}$

C23 Contributed by Chris Black Statement [ 202]
The augmented matrix for the given linear system and its row-reduced form are:

$\begin{array}{l} [\begin{matrix} 1 & - 2 & 1 & - 1 & 0 \\ 1 & 1 & 1 & - 1 & 0 \\ 1 & 0 & 1 & - 1 & 0 \end{matrix}] & \to_{}^{RREF} [\begin{matrix} 1 & 0 & 1 & - 1 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}] . \end{array}$

$\begin{array}{l} [\begin{matrix} - x_{3} + x_{4} \\ 0 \\ x_{3} \\ x_{4} \end{matrix}] \end{array}$

C25 Contributed by Chris Black Statement [ 203]
The augmented matrix for the given linear system and its row-reduced form are:

$\begin{array}{l} [\begin{matrix} 1 & 2 & 3 & 0 \\ 2 & - 1 & 1 & 0 \\ 3 & 1 & 1 & 0 \\ 0 & 1 & 2 & 0 \end{matrix}] & \to_{}^{RREF} [\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}] . \end{array}$

An homogeneous system is always consistent ( Theorem HSC) and with $n = r = 3$ an application of Theorem FVCS yields zero free variables. Thus the only solution to the given system is the trivial solution, $x = 0$ .

C26 Contributed by Chris Black Statement [ 203]
The augmented matrix for the given linear system and its row-reduced form are:

$\begin{array}{l} [\begin{matrix} 1 & 2 & 3 & 0 \\ 2 & - 1 & 1 & 0 \\ 3 & 1 & 1 & 0 \\ 0 & 5 & 2 & 0 \end{matrix}] & \to_{}^{RREF} [\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}] . \end{array}$

C27 Contributed by Chris Black Statement [ 204]
The augmented matrix for the given linear system and its row-reduced form are:

$\begin{array}{l} [\begin{matrix} 1 & 2 & 3 & 0 \\ 2 & - 1 & 1 & 0 \\ 1 & - 8 & - 7 & 0 \\ 0 & 1 & 1 & 0 \end{matrix}] & \to_{}^{RREF} [\begin{matrix} 1 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}] . \end{array}$

An homogeneous system is always consistent ( Theorem HSC) and with $n = 3$ , $r = 2$ an application of Theorem FVCS yields one free variable. With a suitable choice of $x_{3}$ each solution can be written in the form

$\begin{array}{l} [\begin{matrix} - x_{3} \\ - x_{3} \\ x_{3} \end{matrix}] \end{array}$

C30 Contributed by Robert Beezer Statement [ 205]
Definition NSM tells us that the null space of $A$ is the solution set to the homogeneous system $ℒ S (A, 0)$ . The augmented matrix of this system is

$[\begin{matrix} 2 & 4 & 1 & 3 & 8 & 0 \\ - 1 & - 2 & - 1 & - 1 & 1 & 0 \\ 2 & 4 & 0 & - 3 & 4 & 0 \\ 2 & 4 & - 1 & - 7 & 4 & 0 \end{matrix}]$

To solve the system, we row-reduce the augmented matrix and obtain,

$[\begin{matrix} 1 & 2 & 0 & 0 & 5 & 0 \\ 0 & 0 & 1 & 0 & - 8 & 0 \\ 0 & 0 & 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix}]$

This matrix represents a system with equations having three dependent variables ( $x_{1}$ , $x_{3}$ , and $x_{4}$ ) and two independent variables ( $x_{2}$ and $x_{5}$ ). These equations rearrange to

$\begin{array}{l} x_{1} & = - 2 x_{2} - 5 x_{5} & x_{3} & = 8 x_{5} & x_{4} & = - 2 x_{5} \end{array}$

So we can write the solution set (which is the requested null space) as

$N (A) = \{[\begin{matrix} - 2 x_{2} - 5 x_{5} \\ x_{2} \\ 8 x_{5} \\ - 2 x_{5} \\ x_{5} \end{matrix}]| x_{2}, x_{5} \in ℂ^{}\}$

C31 Contributed by Robert Beezer Statement [ 205]
We form the augmented matrix of the homogeneous system $ℒ S (B, 0)$ and row-reduce the matrix,

$\begin{array}{l} [\begin{matrix} - 6 & 4 & - 36 & 6 & 0 \\ 2 & - 1 & 10 & - 1 & 0 \\ - 3 & 2 & - 18 & 3 & 0 \end{matrix}] & \to_{}^{RREF} [\begin{matrix} 1 & 0 & 2 & 1 & 0 \\ 0 & 1 & - 6 & 3 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}] \end{array}$

We knew ahead of time that this system would be consistent ( Theorem HSC), but we can now see there are $n - r = 4 - 2 = 2$ free variables, namely $x_{3}$ and $x_{4}$ (Theorem FVCS). Based on this analysis, we can rearrange the equations associated with each nonzero row of the reduced row-echelon form into an expression for the lone dependent variable as a function of the free variables. We arrive at the solution set to the homogeneous system, which is the null space of the matrix by Definition NSM,

$\begin{array}{l} N (B) = \{[\begin{matrix} - 2 x_{3} - x_{4} \\ 6 x_{3} - 3 x_{4} \\ x_{3} \\ x_{4} \end{matrix}]| x_{3}, x_{4} \in ℂ\} \end{array}$

M50 Contributed by Robert Beezer Statement [ 206]
Since the system is homogeneous, we know it has the trivial solution ( Theorem HSC). We cannot say anymore based on the information provided, except to say that there is either a unique solution or infinitely many solutions (Theorem PSSLS). See Archetype A and Archetype B to understand the possibilities.

M51 Contributed by Robert Beezer Statement [ 206]
Since there are more variables than equations, Theorem HMVEI applies and tells us that the solution set is infinite. From the proof of Theorem HSC we know that the zero vector is one solution.

M52 Contributed by Robert Beezer Statement [ 206]
By Theorem HSC, we know the system is consistent because the zero vector is always a solution of a homogeneous system. There is no more that we can say, since both a unique solution and infinitely many solutions are possibilities.

T10 Contributed by Robert Beezer Statement [ 206]
This is a true statement. A proof is:

( $\Rightarrow$ ) Suppose we have a homogeneous system $ℒ S (A, 0)$ . Then by substituting the scalar zero for each variable, we arrive at true statements for each equation. So the zero vector is a solution. This is the content of Theorem HSC.

( $\Leftarrow$ ) Suppose now that we have a generic (i.e. not necessarily homogeneous) system of equations, $ℒ S (A, b)$ that has the zero vector as a solution. Upon substituting this solution into the system, we discover that each component of $b$ must also be zero. So $b = 0$ .

T20 Contributed by Robert Beezer Statement [ 206]
Suppose that a single equation from this system (the $i$ -th one) has the form,

$a_{i 1} x_{1} + a_{i 2} x_{2} + a_{i 3} x_{3} + \dots + a_{i n} x_{n} = 0$

Evaluate the left-hand side of this equation with the components of the proposed solution vector $v$ ,

$\begin{array}{l} a_{i 1} (4 u_{1}) & + a_{i 2} (4 u_{2}) + a_{i 3} (4 u_{3}) + \dots + a_{i n} (4 u_{n}) \\ = 4 a_{i 1} u_{1} + 4 a_{i 2} u_{2} + 4 a_{i 3} u_{3} + \dots + 4 a_{i n} u_{n} & Commutativity \\ = 4 (a_{i 1} u_{1} + a_{i 2} u_{2} + a_{i 3} u_{3} + \dots + a_{i n} u_{n}) & Distributivity \\ = 4 (0) & u solution to ℒ S (A, 0) \\ = 0 \end{array}$

So $v$ makes each equation true, and so is a solution to the system.

Notice that this result is not true if we change $ℒ S (A, 0)$ from a homogeneous system to a non-homogeneous system. Can you create an example of a (non-homogeneous) system with a solution $u$ such that $v$ is not a solution?