From A First Course in Linear Algebra

Version 1.08

© 2004.

Licensed under the GNU Free Documentation License.

http://linear.ups.edu/

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.

As usual, we begin with a definition.

Definition HS

Homogeneous System

A system of linear equations, $\mathcal{\mathcal{L}}\mathcal{S}\phantom{\rule{0em}{0ex}}\left(A,\phantom{\rule{0em}{0ex}}b\right)$
is homogeneous if the vector of constants is the zero vector, in other words,
$b=0$.
$\u25b3$

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,

Can you quickly find a solution to this system without row-reducing the augmented matrix? $\u22a0$

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. $\square $

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. $\u25a0$

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. $\u25b3$

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,

whose augmented matrix row-reduces to

$$\left[\begin{array}{cccc}\hfill \text{1}\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \text{1}\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \text{1}\hfill & \hfill 0\hfill \end{array}\right]$$ |

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. $\u22a0$

Example HISAA

Homogeneous, infinite solutions, Archetype A

Archetype A can be converted to the homogeneous system,

whose augmented matrix row-reduces to

$$\left[\begin{array}{cccc}\hfill \text{1}\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \text{1}\hfill & \hfill -1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$$ |

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 ${x}_{3}$,

$$S=\left\{\left.\left[\begin{array}{c}\hfill {x}_{1}\hfill \\ \hfill {x}_{2}\hfill \\ \hfill {x}_{3}\hfill \end{array}\right]\phantom{\rule{0em}{0ex}}\right|\phantom{\rule{0em}{0ex}}{x}_{1}=-{x}_{3},\phantom{\rule{0em}{0ex}}{x}_{2}={x}_{3}\right\}=\left\{\left.\left[\begin{array}{c}\hfill -{x}_{3}\hfill \\ \hfill {x}_{3}\hfill \\ \hfill {x}_{3}\hfill \end{array}\right]\phantom{\rule{0em}{0ex}}\right|\phantom{\rule{0em}{0ex}}{x}_{3}\in {\u2102}^{}\right\}$$ |

Geometrically, these are points in three dimensions that lie on a line through the origin. $\u22a0$

Example HISAD

Homogeneous, infinite solutions, Archetype D

Archetype D (and identically, Archetype E) can be converted to the
homogeneous system,

whose augmented matrix row-reduces to

$$\left[\begin{array}{ccccc}\hfill \text{1}\hfill & \hfill 0\hfill & \hfill 3\hfill & \hfill -2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \text{1}\hfill & \hfill 1\hfill & \hfill -3\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$$ |

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 ${x}_{4}$,

$$\begin{array}{llll}\hfill S& =\left\{\left.\left[\begin{array}{c}\hfill {x}_{1}\hfill \\ \hfill {x}_{2}\hfill \\ \hfill {x}_{3}\hfill \\ \hfill {x}_{4}\hfill \end{array}\right]\phantom{\rule{0em}{0ex}}\right|\phantom{\rule{0em}{0ex}}{x}_{1}=-3{x}_{3}+2{x}_{4},\phantom{\rule{0em}{0ex}}{x}_{2}=-{x}_{3}+3{x}_{4}\right\}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left\{\left.\left[\begin{array}{c}\hfill -3{x}_{3}+2{x}_{4}\hfill \\ \hfill -{x}_{3}+3{x}_{4}\hfill \\ \hfill {x}_{3}\hfill \\ \hfill {x}_{4}\hfill \end{array}\right]\phantom{\rule{0em}{0ex}}\right|\phantom{\rule{0em}{0ex}}{x}_{3},\phantom{\rule{0em}{0ex}}{x}_{4}\in {\u2102}^{}\right\}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}& \hfill \end{array}$$ $\u22a0$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.
$\square $

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. $\u25a0$

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).

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 $\mathcal{N}\phantom{\rule{0em}{0ex}}\left(A\right)$,
is the set of all the vectors that are solutions to the homogeneous system
$\mathcal{\mathcal{L}}\mathcal{S}\phantom{\rule{0em}{0ex}}\left(A,\phantom{\rule{0em}{0ex}}0\right)$.

(This definition contains Notation NSM.) $\u25b3$

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.

However, the vector

$$z=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 2\hfill \end{array}\right]$$ |

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. $\u22a0$

Here are two (prototypical) examples of the computation of the null space of a matrix. Notice that we will now begin writing solutions as vectors.

Example CNS1

Computing a null space, #1

Let’s compute the null space of

$$A=\left[\begin{array}{ccccc}\hfill 2\hfill & \hfill -1\hfill & \hfill 7\hfill & \hfill -3\hfill & \hfill -8\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 2\hfill & \hfill 4\hfill & \hfill 9\hfill \\ \hfill 2\hfill & \hfill 2\hfill & \hfill -2\hfill & \hfill -1\hfill & \hfill 8\hfill \end{array}\right]$$ |

which we write as $\mathcal{N}\phantom{\rule{0em}{0ex}}\left(A\right)$. Translating Definition NSM, we simply desire to solve the homogeneous system $\mathcal{\mathcal{L}}\mathcal{S}\phantom{\rule{0em}{0ex}}\left(A,\phantom{\rule{0em}{0ex}}0\right)$. So we row-reduce the augmented matrix to obtain

$$\left[\begin{array}{cccccc}\hfill \text{1}\hfill & \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \text{1}\hfill & \hfill -3\hfill & \hfill 0\hfill & \hfill 4\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill \text{1}\hfill & \hfill 2\hfill & \hfill 0\hfill \end{array}\right]$$ |

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

$$\begin{array}{llll}\hfill {x}_{1}& =-2{x}_{3}-{x}_{5}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {x}_{2}& =3{x}_{3}-4{x}_{5}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {x}_{4}& =-2{x}_{5}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}& \hfill \end{array}$$So we can write the infinite solution set as sets using column vectors,

$$\mathcal{N}\phantom{\rule{0em}{0ex}}\left(A\right)=\left\{\left.\left[\begin{array}{c}\hfill -2{x}_{3}-{x}_{5}\hfill \\ \hfill 3{x}_{3}-4{x}_{5}\hfill \\ \hfill {x}_{3}\hfill \\ \hfill -2{x}_{5}\hfill \\ \hfill {x}_{5}\hfill \end{array}\right]\phantom{\rule{0em}{0ex}}\right|\phantom{\rule{0em}{0ex}}{x}_{3},\phantom{\rule{0em}{0ex}}{x}_{5}\in {\u2102}^{}\right\}$$ |

Example CNS2

Computing a null space, #2

Let’s compute the null space of

$$C=\left[\begin{array}{ccc}\hfill -4\hfill & \hfill 6\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 4\hfill & \hfill 1\hfill \\ \hfill 5\hfill & \hfill 6\hfill & \hfill 7\hfill \\ \hfill 4\hfill & \hfill 7\hfill & \hfill 1\hfill \end{array}\right]$$ |

which we write as $\mathcal{N}\phantom{\rule{0em}{0ex}}\left(C\right)$. Translating Definition NSM, we simply desire to solve the homogeneous system $\mathcal{\mathcal{L}}\mathcal{S}\phantom{\rule{0em}{0ex}}\left(C,\phantom{\rule{0em}{0ex}}0\right)$. So we row-reduce the augmented matrix to obtain

$$\left[\begin{array}{ccc}\hfill \text{1}\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \text{1}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \text{1}\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$$ |

There are no free variables in the homogenous 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

$$\mathcal{N}\phantom{\rule{0em}{0ex}}\left(C\right)=\left\{0\right\}=\left\{\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \end{array}\right]\right\}$$ |

- What is always true of the solution set for a homogeneous system of equations?
- Suppose a homogeneous sytem of equations has 13 variables and 8 equations. How many solutions will it have? Why?
- Describe in words (not symbols) the null space of a matrix.

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

C30 Compute the null space of the matrix $A$, $\mathcal{N}\phantom{\rule{0em}{0ex}}\left(A\right)$.

$$A=\left[\begin{array}{ccccc}\hfill 2\hfill & \hfill 4\hfill & \hfill 1\hfill & \hfill 3\hfill & \hfill 8\hfill \\ \hfill -1\hfill & \hfill -2\hfill & \hfill -1\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 4\hfill & \hfill 0\hfill & \hfill -3\hfill & \hfill 4\hfill \\ \hfill 2\hfill & \hfill 4\hfill & \hfill -1\hfill & \hfill -7\hfill & \hfill 4\hfill \end{array}\right]$$ |

Contributed by Robert Beezer Solution [170]

C31 Find the null space of the matrix $B$, $\mathcal{N}\phantom{\rule{0em}{0ex}}\left(B\right)$.

$$\begin{array}{llll}\hfill B& =\left[\begin{array}{cccc}\hfill -6\hfill & \hfill 4\hfill & \hfill -36\hfill & \hfill 6\hfill \\ \hfill 2\hfill & \hfill -1\hfill & \hfill 10\hfill & \hfill -1\hfill \\ \hfill -3\hfill & \hfill 2\hfill & \hfill -18\hfill & \hfill 3\hfill \end{array}\right]\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$

Contributed by Robert Beezer Solution [172]

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 [173]

M51 A homogeneous system of 8 equations in 9 variables.

Contributed by Robert Beezer Solution [173]

M52 A homogeneous system of 8 equations in 7 variables.

Contributed by Robert Beezer Solution [173]

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 [173]

T20 Consider the homogeneous system of linear equations
$\mathcal{\mathcal{L}}\mathcal{S}\phantom{\rule{0em}{0ex}}\left(A,\phantom{\rule{0em}{0ex}}0\right)$, and suppose
that $u=\left[\begin{array}{c}\hfill {u}_{1}\hfill \\ \hfill {u}_{2}\hfill \\ \hfill {u}_{3}\hfill \\ \hfill \vdots \hfill \\ \hfill {u}_{n}\hfill \end{array}\right]$
is one solution to the system of equations. Prove that
$v=\left[\begin{array}{c}\hfill 4{u}_{1}\hfill \\ \hfill 4{u}_{2}\hfill \\ \hfill 4{u}_{3}\hfill \\ \hfill \vdots \hfill \\ \hfill 4{u}_{n}\hfill \end{array}\right]$is also a
solution to $\mathcal{\mathcal{L}}\mathcal{S}\phantom{\rule{0em}{0ex}}\left(A,\phantom{\rule{0em}{0ex}}0\right)$.

Contributed by Robert Beezer Solution [174]

C30 Contributed by Robert Beezer Statement [166]

Definition NSM tells us that the null space of
$A$ is the solution set to the
homogeneous system $\mathcal{\mathcal{L}}\mathcal{S}\phantom{\rule{0em}{0ex}}\left(A,\phantom{\rule{0em}{0ex}}0\right)$.
The augmented matrix of this system is

$$\left[\begin{array}{cccccc}\hfill 2\hfill & \hfill 4\hfill & \hfill 1\hfill & \hfill 3\hfill & \hfill 8\hfill & \hfill 0\hfill \\ \hfill -1\hfill & \hfill -2\hfill & \hfill -1\hfill & \hfill -1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill 4\hfill & \hfill 0\hfill & \hfill -3\hfill & \hfill 4\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill 4\hfill & \hfill -1\hfill & \hfill -7\hfill & \hfill 4\hfill & \hfill 0\hfill \end{array}\right]$$ |

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

$$\left[\begin{array}{cccccc}\hfill \text{1}\hfill & \hfill 2\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 5\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \text{1}\hfill & \hfill 0\hfill & \hfill -8\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill \text{1}\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$$ |

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

$$\begin{array}{llllllllllll}\hfill {x}_{1}& =-2{x}_{2}-5{x}_{5}\phantom{\rule{2em}{0ex}}& \hfill {x}_{3}& =8{x}_{5}\phantom{\rule{2em}{0ex}}& \hfill {x}_{4}& =-2{x}_{5}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$So we can write the solution set (which is the requested null space) as

$$\mathcal{N}\phantom{\rule{0em}{0ex}}\left(A\right)=\left\{\left.\left[\begin{array}{c}\hfill -2{x}_{2}-5{x}_{5}\hfill \\ \hfill {x}_{2}\hfill \\ \hfill 8{x}_{5}\hfill \\ \hfill -2{x}_{5}\hfill \\ \hfill {x}_{5}\hfill \end{array}\right]\phantom{\rule{0em}{0ex}}\right|\phantom{\rule{0em}{0ex}}{x}_{2},{x}_{5}\in {\u2102}^{}\right\}$$ |

C31 Contributed by Robert Beezer Statement [167]

We form the augmented matrix of the homogeneous system
$\mathcal{\mathcal{L}}\mathcal{S}\phantom{\rule{0em}{0ex}}\left(B,\phantom{\rule{0em}{0ex}}0\right)$ and
row-reduce the matrix,

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,

$$\begin{array}{lll}\hfill \mathcal{N}\phantom{\rule{0em}{0ex}}\left(B\right)=\left\{\left.\left[\begin{array}{c}\hfill -2{x}_{3}-{x}_{4}\hfill \\ \hfill 6{x}_{3}-3{x}_{4}\hfill \\ \hfill {x}_{3}\hfill \\ \hfill {x}_{4}\hfill \end{array}\right]\phantom{\rule{0em}{0ex}}\right|\phantom{\rule{0em}{0ex}}{x}_{3},\phantom{\rule{0em}{0ex}}{x}_{4}\in \u2102\right\}& \phantom{\rule{2em}{0ex}}& \hfill \end{array}$$M50 Contributed by Robert Beezer Statement [168]

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 [168]

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 [168]

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 [168]

This is a true statement. A proof is:

($\Rightarrow $) Suppose we have a homogeneous system $\mathcal{\mathcal{L}}\mathcal{S}\phantom{\rule{0em}{0ex}}\left(A,\phantom{\rule{0em}{0ex}}0\right)$. 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, $\mathcal{\mathcal{L}}\mathcal{S}\phantom{\rule{0em}{0ex}}\left(A,\phantom{\rule{0em}{0ex}}b\right)$ 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 [168]

Suppose that a single equation from this system (the
$i$-th
one) has the form,

$${a}_{i1}{x}_{1}+{a}_{i2}{x}_{2}+{a}_{i3}{x}_{3}+\cdots +{a}_{in}{x}_{n}=0$$ |

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

$$\begin{array}{llllll}\hfill {a}_{i1}\left(4{u}_{1}\right)& +{a}_{i2}\left(4{u}_{2}\right)+{a}_{i3}\left(4{u}_{3}\right)+\cdots +{a}_{in}\left(4{u}_{n}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =4{a}_{i1}{u}_{1}+4{a}_{i2}{u}_{2}+4{a}_{i3}{u}_{3}+\cdots +4{a}_{in}{u}_{n}\phantom{\rule{2em}{0ex}}& \hfill & \text{Commutativity}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =4\left({a}_{i1}{u}_{1}+{a}_{i2}{u}_{2}+{a}_{i3}{u}_{3}+\cdots +{a}_{in}{u}_{n}\right)\phantom{\rule{2em}{0ex}}& \hfill & \text{Distributivity}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =4\left(0\right)\phantom{\rule{2em}{0ex}}& \hfill & \text{}u\text{solutionto}\mathcal{\mathcal{L}}\mathcal{S}\phantom{\rule{0em}{0ex}}\left(A,\phantom{\rule{0em}{0ex}}0\right)\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$So $v$ makes each equation true, and so is a solution to the system.

Notice that this result is not true if we change $\mathcal{\mathcal{L}}\mathcal{S}\phantom{\rule{0em}{0ex}}\left(A,\phantom{\rule{0em}{0ex}}0\right)$ 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?