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Section HSE Homogeneous 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.

Subsection SHS Solutions of Homogeneous Systems

As usual, we begin with a definition.

Definition HS. Homogeneous 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 \(\vect{b}=\zerovector\text{.}\)

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{align*} 2x_1 - 3x_2 + x_3 - 6x_4 &= 0 \\ 4x_1 +x_2 +2x_3 + 9x_4 &= 0 \\ 3x_1 +x_2 +x_3 + 8x_4 &= 0\text{.} \end{align*}

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.

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 \(x_1=0\text{,}\) \(x_2=0\text{,}\) …, \(x_n=0\) (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.

Archetype B can be converted to the homogeneous system,

\begin{align*} -7x_1 -6 x_2 - 12x_3 &=0\\ 5x_1 + 5x_2 + 7x_3 &=0\\ x_1 +4x_3 &=0 \end{align*}

whose augmented matrix row-reduces to

\begin{equation*} \begin{bmatrix} \leading{1} & 0 & 0 & 0\\ 0 & \leading{1} & 0 & 0\\ 0 & 0 & \leading{1} & 0 \end{bmatrix}\text{.} \end{equation*}

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.

Archetype A can be converted to the homogeneous system

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

whose augmented matrix row-reduces to

\begin{equation*} \begin{bmatrix} \leading{1} & 0 & 1 & 0 \\ 0 & \leading{1} & -1 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}\text{.} \end{equation*}

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 \(x_3\text{.}\)

\begin{equation*} S=\setparts{\colvector{x_1\\x_2\\x_3}}{x_1=-x_3,\,x_2=x_3}=\setparts{\colvector{-x_3\\x_3\\x_3}}{x_3\in\complexes} \end{equation*}

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

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

\begin{align*} 2x_1 + x_2 + 7x_3 - 7x_4 &= 0 \\ -3x_1 + 4x_2 -5x_3 - 6x_4 &= 0 \\ x_1 +x_2 + 4x_3 - 5x_4 &= 0 \end{align*}

whose augmented matrix row-reduces to

\begin{equation*} \begin{bmatrix} \leading{1} & 0 & 3 & -2 & 0 \\ 0 & \leading{1} & 1 & -3 & 0\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}\text{.} \end{equation*}

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 \(x_4\text{.}\)

\begin{align*} S&=\setparts{\colvector{x_1\\x_2\\x_3\\x_4}}{x_1=-3x_3+2x_4,\,x_2=-x_3+3x_4}\\ &=\setparts{\colvector{-3x_3+2x_4\\-x_3+3x_4\\x_3\\x_4}}{ x_3,\,x_4\in\complexes} \end{align*}

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.

We are assuming the system is homogeneous, so Theorem HSC says it is consistent. Then the hypothesis that \(n\gt m\text{,}\) 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 \(n\gt m\) where Theorem HMVEI tells us that there is only one possibility for a homogeneous system).

Sage SHS. Solving Homogeneous Systems.

We can explore homogeneous systems easily in Sage with commands we have already learned. Notably, the zero_vector()constructor will quickly create the necessary vector of constants (Sage V).

You could try defining a function to accept a matrix as input, augment the matrix with a zero vector having the right number of entries, and then return the reduced row-echelon form of the augmented matrix, or maybe you could return the number of free variables. (See Sage RCLS for a refresher on how to do this). It is also interesting to see how .solve_right()behaves on homogeneous systems, since in particular we know it will never give us an error. (Why not? Hint: Theorem HSC.)

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 \(A\text{,}\) 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.

The write-up for Archetype I lists several solutions of the corresponding homogeneous system. The vectors \(\vect{x}\) and \(\vect{y}\) 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 \(\vect{z}\) 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.

\begin{align*} \vect{x}=\colvector{3\\0\\-5\\-6\\0\\0\\1}&&\vect{y}=\colvector{-4\\1\\-3\\-2\\1\\1\\1}&& \vect{z}=\colvector{1\\0\\0\\0\\0\\0\\2}& \end{align*}

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

Let us compute the null space of

\begin{equation*} A=\begin{bmatrix} 2 & -1 & 7 & -3 & -8 \\ 1 & 0 & 2 & 4 & 9 \\ 2 & 2 & -2 & -1 & 8 \end{bmatrix} \end{equation*}

which we write as \(\nsp{A}\text{.}\) Translating Definition NSM, we simply desire to solve the homogeneous system \(\homosystem{A}\text{.}\) So we row-reduce the augmented matrix to obtain

\begin{equation*} \begin{bmatrix} \leading{1} & 0 & 2 & 0 & 1 & 0 \\ 0 & \leading{1} & -3 & 0 & 4 & 0 \\ 0 & 0 & 0 & \leading{1} & 2 & 0 \end{bmatrix}\text{.} \end{equation*}

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{align*} x_1&=-2x_3-x_5\\ x_2&=3x_3-4x_5\\ x_4&=-2x_5\text{.} \end{align*}

So we can write the infinite solution set using column vectors.

\begin{equation*} \nsp{A}=\setparts{\colvector{-2x_3-x_5\\3x_3-4x_5\\x_3\\-2x_5\\x_5} }{x_3,\,x_5\in\complexes} \end{equation*}

Let us compute the null space of

\begin{equation*} C=\begin{bmatrix} -4 & 6 & 1 \\ -1 & 4 & 1 \\ 5 & 6 & 7 \\ 4 & 7 & 1 \end{bmatrix} \end{equation*}

which we write as \(\nsp{C}\text{.}\) Translating Definition NSM, we simply desire to solve the homogeneous system \(\homosystem{C}\text{.}\) So we row-reduce the augmented matrix to obtain

\begin{equation*} \begin{bmatrix} \leading{1} & 0 & 0 & 0 \\ 0 & \leading{1} & 0 & 0 \\ 0 & 0 & \leading{1} & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}\text{.} \end{equation*}

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, \(\zerovector\text{.}\) So we can write the solution set as

\begin{equation*} \nsp{C}=\set{\zerovector}=\set{\colvector{0\\0\\0}}\text{.} \end{equation*}

Sage NS. Null Space.

Sage will compute a null space for us. Which is rather remarkable, as it is an infinite set! Again, this is a powerful command, and there is lots of associated theory, so we will not understand everything about it right away, and it also has a radically different name in Sage. But we will find it useful immediately. Let us reprise Example¬†NSEAI. The relevant command to build the null space of a matrix is .right_kernel(), where again, we will rely exclusively on the ‚Äúright‚ÄĚ version. Also, to match our work in the text, and make the results more recognizable, we will consistently use the keyword option basis='pivot', which we will be able to explain once we have more theory (Sage¬†SSNS, Sage¬†SUTH0). Note too, that this is a place where it is critical that matrices are defined to use the rationals as their number system (QQ).

As we said, nspcontains a lot of unfamiliar information. Ignore most of it for now. But as a set, we can test membership in nsp.

We did a bad thing above, as Sage likes to use Ifor the imaginary number \(i=\sqrt{-1}\) and we just clobbered that. We will not do it again. See below how to fix this.

nspis an infinite set. Since we know the null space is defined as solutions to a system of equations, and the work above shows it has at least two elements, we are not surprised to discover that the set is infinite (Theorem PSSLS).

If we want an element of the null space to experiment with, we can get a ‚Äúrandom‚ÄĚ element easily. Evaluate the following compute cell repeatedly to get a feel for the variety of the different output. You will see a different result each time, and the result supplied here originally is very unlikely to be a result you will ever see again. The bit of text, # random, is technically a ‚Äúcomment‚ÄĚ, but we are using it as a signal to our automatic testing of the Sage examples that this example should be skipped. You do not need to use this device in your own work, though you may use the comment syntax if you wish.

Sometimes, just sometimes, the null space is finite, and we can list its elements. This is from Example CNS2.

Notice that we get back a list (which mathematically is really a set), and it has one element, the three-entry zero vector.

There is more to learn about exploring the null space with Sage's .right_kernel()so we will see more of this matrix method. In the meantime, if you are done experimenting with the matrix Iwe can restore the variable Iback to being the imaginary number \(i = \sqrt{-1}\) with the Sage restore()command.

Sage SH. Sage Help.

There are many ways to learn about, or remind yourself of, how various Sage commands behave. Now that we have learned a few, it is a good time to show you the most direct methods of obtaining help. These work throughout Sage, so can be useful if you want to apply Sage to other areas of mathematics.

The first hurdle is to learn how to make a mathematical object in Sage. We now know how to make matrices and vectors (and null spaces). This is enough to help us explore relevant commands in Sage for linear algebra. First, define a very simple matrix A, with maybe one row and two columns. The number system you choose will have some effect on the results, so use QQfor now. In a cell enter A.(assuming you called your matrix A, and be sure to include the period). Now hit the ‚ÄúTAB‚ÄĚ key and you will get a long list of all the possible methods you can apply to Ausing the dot notation.

You can click directly on one of these commands to enter it into the cell. Now instead of adding parentheses to the command, place a single question mark (?) on the end and hit the TABkey again. You should get some nicely formatted documentation, along with example uses. (Try A.rref?below for a good example of this.) You can replace the single question mark by two question marks, and as Sage is an open source program you can see the actual computer instructions for the method, which at first includes all the documentation again. Note that now the documentation is enclosed in a pair of triple quotation marks (""", """) as part of the source code, and is not specially formatted. (Notice that the use of TABwill not work as advertised with the Sage Cell server.)

These methods of learning about Sage are generally referred to as tab-completion and we will use this term going forward. To learn about the use of Sage in other areas of mathematics, you just need to find out how to create the relevant objects via a ‚Äúconstructor‚ÄĚ function, such as matrix()and vector()for linear algebra. Sage has a comprehensive Reference Manual and there is a Linear Algebra Quick Reference sheet. These should be easily located online via sagemath.org‚ÄČ1‚ÄČ or with an internet search leading with the terms ‚Äúsage math‚ÄĚ (use ‚Äúmath‚ÄĚ to avoid confusion with other websites for things named ‚ÄúSage‚ÄĚ).

Reading Questions HSE Reading 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.)

Exercises HSE 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, 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.

Exercise Group.

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.

The augmented matrix for the given linear system and its row-reduced form are:

\begin{align*} \begin{bmatrix} 1 & 4 & 3 & -1 & 0\\ 1 & -1 & 1 & 2 & 0\\ 4 & 1 & 6 & 5 & 0 \end{bmatrix} &\rref \begin{bmatrix} \leading{1} & 0 & 7/5 & 7/5 & 0\\ 0 & \leading{1} & 2/5 & -3/5 & 0\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}\text{.} \end{align*}

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 \(x_4\text{,}\) each solution can be written as

\begin{gather*} \colvector{-\frac{7}{5} x_3 - \frac{7}{5} x_4 \\ -\frac{2}{5} x_3 + \frac{3}{5} x_4 \\ x_3 \\ x_4}\text{.} \end{gather*}
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.

The augmented matrix for the given linear system and its row-reduced form are:

\begin{align*} \begin{bmatrix} 1 & -2 & 1 & -1 & 0\\ 2 & -4 & 1 & 1 & 0\\ 1 & -2 & -2 & 3 & 0 \end{bmatrix} &\rref \begin{bmatrix} \leading{1} & -2 & 0 & 0 & 0\\ 0 & 0 & \leading{1} & 0 & 0\\ 0 & 0 & 0 & \leading{1} & 0 \end{bmatrix}\text{.} \end{align*}

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 \(x_2\text{,}\) each solution can be written as

\begin{gather*} \colvector{2x_2\\ x_2 \\ 0 \\ 0}\text{.} \end{gather*}
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.

The augmented matrix for the given linear system and its row-reduced form are:

\begin{align*} \begin{bmatrix} 1 & -2 & 1 & -1 & 0\\ 1 & 1 & 1 & -1 & 0\\ 1 & 0 & 1 & -1 & 0 \end{bmatrix} &\rref \begin{bmatrix} \leading{1} & 0 & 1 & -1 & 0\\ 0 & \leading{1} & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}\text{.} \end{align*}

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 \(x_4\text{,}\) each solution can be written as

\begin{gather*} \colvector{-x_3 + x_4 \\ 0 \\ x_3 \\ x_4}\text{.} \end{gather*}

Exercise Group.

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.

The augmented matrix for the given linear system and its row-reduced form are:

\begin{align*} \begin{bmatrix} 1 & 2 & 3 & 0\\ 2 & -1 & 1 & 0\\ 3 & 1 & 1 & 0\\ 0 & 1 & 2 & 0 \end{bmatrix} &\rref \begin{bmatrix} \leading{1} & 0 & 0 & 0\\ 0 & \leading{1} & 0 & 0\\ 0 & 0 & \leading{1} & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}\text{.} \end{align*}

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, \(\vect{x} = \zerovector\text{.}\)

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.

The augmented matrix for the given linear system and its row-reduced form are:

\begin{align*} \begin{bmatrix} 1 & 2 & 3 & 0\\ 2 & -1 & 1 & 0\\ 3 & 1 & 1 & 0\\ 0 & 5 & 2 & 0 \end{bmatrix} &\rref \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}\text{.} \end{align*}

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, \(\vect{x} = \zerovector\text{.}\)

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.

The augmented matrix for the given linear system and its row-reduced form are:

\begin{align*} \begin{bmatrix} 1 & 2 & 3 & 0\\ 2 & -1 & 1 & 0\\ 1 & -8 & -7 & 0\\ 0 & 1 & 1 & 0 \end{bmatrix} &\rref \begin{bmatrix} \leading{1} & 0 & 1 & 0\\ 0 & \leading{1} & 1 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}\text{.} \end{align*}

An homogeneous system is always consistent (Theorem HSC) and with \(n=3\text{,}\) \(r=2\) 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{gather*} \colvector{-x_3 \\ -x_3 \\ x_3}\text{.} \end{gather*}

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.

Definition NSM tells us that the null space of \(A\) is the solution set to the homogeneous system \(\homosystem{A}\text{.}\) The augmented matrix of this system is

\begin{align*} \begin{bmatrix} 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{bmatrix}\text{.} \end{align*}

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

\begin{equation*} \begin{bmatrix} \leading{1} & 2 & 0 & 0 & 5 & 0 \\ 0 & 0 & \leading{1} & 0 & -8 & 0 \\ 0 & 0 & 0 & \leading{1} & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}\text{.} \end{equation*}

This matrix represents a system with equations having three dependent variables (\(x_1\text{,}\) \(x_3\text{,}\) and \(x_4\)) and two independent variables (\(x_2\) and \(x_5\)). These equations rearrange to

\begin{align*} x_1&=-2x_2-5x_5 & x_3&=8x_5 & x_4&=-2x_5\text{.} \end{align*}

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

\begin{equation*} \nsp{A}=\setparts{\colvector{-2x_2-5x_5\\x_2\\8x_5\\-2x_5\\x_5} }{x_2, x_5\in\complexes }\text{.} \end{equation*}

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.

We form the augmented matrix of the homogeneous system \(\linearsystem{B}{\zerovector}\) and row-reduce the matrix,

\begin{align*} \begin{bmatrix} -6 & 4 & -36 & 6 & 0 \\ 2 & -1 & 10 & -1 & 0 \\ -3 & 2 & -18 & 3 & 0 \end{bmatrix} &\rref \begin{bmatrix} \leading{1} & 0 & 2 & 1 & 0 \\ 0 & \leading{1} & -6 & 3 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}\text{.} \end{align*}

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{gather*} \nsp{B}=\setparts{\colvector{-2x_3-x_4\\6x_3-3x_4\\x_3\\x_4}}{x_3,\,x_4\in\complexes} \end{gather*}

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

Exercise Group.

For Exercises M50‚ÄďM52 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.

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.

A homogeneous system of 8 equations in 9 variables.

Solution.

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.

A homogeneous system of 8 equations in 7 variables.

Solution.

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.

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

Solution.

This is a true statement. A proof is:

Suppose we have a homogeneous system \(\homosystem{A}\text{.}\) 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.

‚áź Suppose now that we have a generic (i.e. not necessarily homogeneous) system of equations, \(\linearsystem{A}{\vect{b}}\) that has the zero vector as a solution. Upon substituting this solution into the system, we discover that each component of \(\vect{b}\) must also be zero. So \(\vect{b}=\zerovector\text{.}\)

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.

If the first system is homogeneous, then the zero vector is in the solution set of the system. (See the proof of Theorem HSC.)

Since the two systems are equivalent, they have equal solutions sets (Definition ESYS). So the zero vector is in the solution set of the second system. By Exercise HSE.T10 the presence of the zero vector in the solution set implies that the system is homogeneous.

So if any one of two equivalent systems is homogeneous, then they both are homogeneous.

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.

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

\begin{equation*} a_{i1}x_1+a_{i2}x_2+a_{i3}x_3+\dots+a_{in}x_n=0\text{.} \end{equation*}

Evaluate the left-hand side of this equation with the components of the proposed solution vector \(\vect{v}\)

\begin{align*} a_{i1}\left(4u_1\right)&+a_{i2}\left(4u_2\right)+a_{i3}\left(4u_3\right)+\dots+a_{in}\left(4u_n\right)\\ &=4a_{i1}u_1+4a_{i2}u_2+4a_{i3}u_3+\dots+4a_{in}u_n&& \text{Commutativity}\\ &=4\left(a_{i1}u_1+a_{i2}u_2+a_{i3}u_3+\dots+a_{in}u_n\right)&& \text{Distributivity}\\ &=4(0)&& \vect{u}\text{ solution to }\homosystem{A}\\ &=0\text{.} \end{align*}

So \(\vect{v}\) makes each equation true, and so is a solution to the system.

Notice that this result is not true if we change \(\homosystem{A}\) from a homogeneous system to a non-homogeneous system. Can you create an example of a (non-homogeneous) system with a solution \(\vect{u}\) such that \(\vect{v}\) is not a solution?