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Linear algebra: the theory of vector spaces and linear transformations

Aaron Greicius

Section 1.1 Systems of linear equations

Computing and studying solutions to equations, and systems of equations, unquestionably plays an important role in mathematics; though we hasten to add that this is not all that mathematicians do! In this chapter we will develop an essentially complete theory of a particularly simple family of mathematical equations: namely, linear equations. This will serve as a somewhat indirect introduction to our study of linear algebra, as lurking below our parametric descriptions of solutions to linear systems lie vector space notions like subspace, span, and linear independence. Furthermore, we will meet one of the most important computational tools of linear algebra: Gaussian elimination.

Subsection 1.1.1 Systems of linear equations

Definition 1.1.1. Linear equations.

A linear expression in the \(n\) unknowns (or variables) \(x_1,x_2,\dots x_n\) is an expression of the form
\begin{equation*} a_1x_1+a_2x_2+\cdots +a_nx_n\text{,} \end{equation*}
where \(a_1,a_2,\dots, a_n\) are fixed real numbers.
A linear equation in the unknowns \(x_1,x_2,\dots, x_n\) is an equation that can be simplified, using only addition and subtraction, to an equation of the form
\begin{equation} a_1x_1+a_2x_2+\cdots a_nx_n=b,\tag{1.1.1} \end{equation}
which we call its standard form. An equation in the unknowns \(x_1,x_2,\dots, x_n\) is nonlinear if it cannot be simplified to the form (1.1.1) using only addition and subtraction.
Given a linear equation with standard form (1.1.1), the equation is called homogeneous if \(b=0\text{,}\) and nonhomogeneous if \(b\ne 0\text{.}\)

Example 1.1.2. Linear and nonlinear equations.

  1. Consider \(\sqrt{3}x+\sin(5)=2z-e^4y\text{.}\) This is a linear equation in the unknowns \(x,y,z\text{.}\) Its standard form is \(\sqrt{3}x+e^4y-2z=-\sin(5)\text{.}\) Since the right-hand side is nonzero, we see that the equation is nonhomogeneous.
  2. The equation \(x^2+y^2=1\) is a nonlinear equation in the unknowns \(x\) and \(y\text{.}\)

Definition 1.1.3. Systems of linear equations.

A system of linear equations (or linear system) is a set of linear equations.
A homogeneous linear system is a set of homogeneous linear equations.
When displaying a system of \(m\) equations in the \(n\) unknowns \(x_1,x_2,\dots x_n\text{,}\) we typically write each equation in standard form and align the corresponding terms of each equation into columns:
\begin{equation*} \eqsys \end{equation*}
A homogeneous system is thus typically written as:
\begin{equation*} \homsys. \end{equation*}

Remark 1.1.4.

You will want to get comfortable with the double-indexing used to display linear systems as quickly as possible. Here is a good way to orient yourself:
  • The \(i\) appearing in \(a_{ij}\) and \(b_i\) indicates the \(i\)-th row in our displayed system, or equivalently, the \(i\)-th equation.
  • The \(j\) appearing in \(a_{ij}\) indicates the \(j\)-th column in our displayed system, which is associated to the \(j\)-th variable, for \(1\leq j\leq n\text{.}\)

Definition 1.1.5. Solutions to linear systems.

A solution to a linear equation
\begin{equation*} a_1x_1+a_2x_2+\cdots a_nx_n=b \end{equation*}
is an \(n\)-tuple \((s_1,s_2,\dots, s_n)\) of real numbers for which the variable assignment \(x_1=s_1, x_2=s_2,\dots, x_n=s_n\) makes the equation true. We say \((s_1,\dots ,s_n)\) solves the equation in this case.
A solution to a system of linear equations
\begin{equation*} \eqsys \end{equation*}
is an \(n\)-tuple \((s_1,s_2,\dots,s_n)\) that is a solution to each of the system’s \(m\) equations. We say \((s_1,s_2,\dots, s_n)\) solves the system in this case.
Given a linear system we will seek to find the set of all solutions to the system. As we will soon see, this set of solutions will take one of three qualitative forms:
  1. The solution set is empty; i.e., there are no solutions. We say the system is inconsistent in this case. Otherwise a system is called consistent.
  2. The solution set contains a single element; i.e., there is exactly one solution.
  3. The solution set contains infinitely many elements; i.e., there are infinitely solutions.

Example 1.1.6. Solutions to elementary systems.

For each system, compute the set of solutions.
  1. \begin{equation*} \begin{linsys}{2} x\amp -\amp y\amp =\amp 0\\ x\amp -\amp y\amp =\amp 1\end{linsys} \end{equation*}
  2. \begin{equation*} \begin{linsys}{2} x\amp -\amp y\amp =\amp 0\\ x\amp +\amp y\amp =\amp 1\end{linsys} \end{equation*}
  3. \begin{equation*} \begin{linsys}{2} x\amp -\amp y\amp =\amp 1\\ 2x\amp -\amp 2y\amp =\amp 2\end{linsys} \end{equation*}
Solution.
  1. The first equation implies \(x=y\text{.}\) Substituting \(y\) for \(x\) in the second equation, we would then have \(0=1\text{,}\) a contradiction. Thus there are no solutions: i.e., \(S\) is the empty set, denoted \(S=\{ \ \}\) or \(S=\emptyset\text{.}\)
  2. The first equation implies \(x=y\text{.}\) Making this substitution in the second equation yields \(2x=1\text{,}\) or \(x=1/2\text{.}\) Thus \((x,y)=(1/2,1/2)\) is the unique solution, and \(S=\{ (1/2,1/2)\}\text{.}\)
  3. The second equation is just twice the first. It follows that both equations have the exact same set of solutions, and so we need only find all solutions to \(x-y=1\text{.}\) Note that we can set \(x=t\) for any real number \(t \in\R\text{.}\) Solving for \(y\) in terms of \(t\) we get \((x,y)=(t,t-1)\) for any \(t\in\R\text{,}\) and thus \(S=\{(t,t-1)\colon t\in\R\}\text{,}\) an infinite set!
As you may recall, a (nontrivial) linear equation in two unknowns defines a line in \(\R^2\text{;}\) and a (nontrivial) equation in three unknowns defines a plane in \(\R^3\text{.}\) This observation allows us to bring powerful geometric intuition to the analysis of systems of linear equations in two or three unknowns. As lines and planes will also serve as a rich source of examples in this course, we recall below some of their basic notions.

Example 1.1.7. Lines and planes review.

Below we define lines (in \(\R^2\)) and planes (in \(\R^3\)) as sets of solutions of linear equations in two or three unknowns. Recall that the graph of an equation is the set of all solutions to the equation.
  • Lines.
    A line in \(\R^2\) is the set of solutions (or graph) \(\ell\) of a linear equation of the form \(ax+by=c\text{,}\) where \(a, b, c\in\R\) are fixed constants, and at least one of \(a\) and \(b\) is nonzero: i.e.,
    \begin{equation*} \ell=\{(x,y)\in\R^2\colon ax+by=c\}. \end{equation*}
  • Planes.
    A plane in \(\R^3\) is the set of solutions (or graph) \(\mathcal{P}\) of a linear equation of the form \(ax+by+cz=d\text{,}\) where \(a, b, c, d\in\R\) are fixed constants, and at least one of \(a, b, c\) is nonzero: i.e.,
    \begin{equation*} \mathcal{P}=\{(x,y,z)\in\R^3\colon ax+by+cz=d\}. \end{equation*}
We often use the abbreviated notation
\begin{align*} \ell \amp\colon ax+by=c \amp \mathcal{P}\amp\colon ax+by+cz=d \end{align*}
to introduce the line \(\ell\) with defining equation \(ax+by=c\) or plane \(\mathcal{P}\) with defining equation \(ax+by+cz=d\text{.}\)
Given a plane \(\mathcal{P}\colon ax+by+cz=d\text{,}\) we call \(\boldn=(a,b,c)\) a normal vector of \(\mathcal{P}\text{.}\) Equipped with the normal vector \(\boldn\) and a single point \(P=(x_0,y_0,z_0)\) lying on \(\mathcal{P}\text{,}\) we can quickly visualize \(\mathcal{P}\) as the set of all points \(Q=(x,y,z)\) such that the vector \(\overrightarrow{PQ}=(x-x_0, y-y_0, z-z_0)\) is orthogonal to \(\mathcal{n}\text{.}\) This allows us to envision \(\mathcal{P}\) as “the plane passing through \(P\) and orthogonal to \(\boldn\)”. The customizable Sage Cell below nicely demonstrates this method of visualization.
Figure 1.1.8. Using Sage to visualize \(\mathcal{P}\colon ax+by+cz=d\) via normal vector \(\boldn=(a,b,c)\)

Example 1.1.9. Solutions to elementary systems (again).

For each system in Example 1.1.6, use a geometric argument to determine how many solutions there are.
Solution.
We proceed geometrically by observing that each equation in a given system defines a line, that a pair \((x,y)\) is a solution to the given equation if and only if it is a point on the corresponding line, and thus that a pair \((x,y)\) is a solution to both equations if and only if it is a point lying on both lines. Thus, geometrically speaking, the set of solutions to a given system is equal to the intersection of the two corresponding lines.
  1. In this case the two lines \(\ell_1\colon x-y=0\) and \(\ell_2\colon x-y=1\) are parallel (and not equal to one another). It follows that their intersection is empty. We conclude that there are no solutions to the system.
  2. In this case the two lines \(\ell_1\colon x-y=0\) and \(\ell_2\colon x+y=1\) are nonparallel. It follows that they intersect in a single point. We conclude that the system has a unique solution.
  3. In this case the two equations \(x-y=1\) and \(2x-2y=2\) define the same line \(\ell\text{,}\) and thus that their intersection is also equal to \(\ell\text{.}\) The set of solutions to the system is then equal to the set of points of \(\ell\text{,}\) which is infinite.

Subsection 1.1.2 Row operations

In Example 1.1.6 we were able to combine some simple logic and arithmetic to solve each system completely. Of course, things get more complicated as the number of equations and/or unknowns in the system increases.
For example, consider the system
\begin{equation} \begin{linsys}{3} 2x\amp -\amp y\amp -\amp z\amp =\amp 3\\ x\amp \amp \amp -\amp z\amp =\amp 2 \\ x\amp -\amp y\amp \amp \amp =\amp 1 \end{linsys}\tag{1.1.2} \end{equation}
Observe first that the triple \((5,2,5)\) is not a solution to the system; one easily checks that is satisfies the first and second equations, but not the third. Similarly, it is easy to verify that \((4,2,3)\) and \((0,-1,2)\) are both solutions to the system. The question remains: How do we find all solutions to the linear system in a systematic way?

Remark 1.1.10.

Notice the somewhat funny spacing in (1.1.2). This is in force in order to align the unknowns in separate columns. In general, when handed a linear system in the wild, your first step should be to write each equation in standard form, and make sure the unknowns are aligned vertically in this manner.

Some systems are easier to solve than others.

Below you find two systems of three equations in three unknowns.
\begin{equation*} \begin{array}{ccc} \text{ System } L \amp \hspace{5pt}\amp \text{ System } L' \\ \hline \begin{linsys}{3} 2x\amp +\amp 3y\amp +\amp -z\amp =\amp 18\\ x\amp +\amp 2y\amp -\amp 2z\amp =\amp 8\\ -\frac{1}{2}x\amp +\amp -\frac{1}{2}y\amp +\amp \frac{1}{2}z\amp =\amp -3 \end{linsys} \amp \amp \begin{linsys}{3} x\amp +\amp y\amp +\amp z\amp =\amp 10\\ \amp \amp y\amp -\amp 3z\amp =\amp -2\\ \amp \amp \amp \amp z\amp =\amp 2 \end{linsys} \end{array} \end{equation*}
The staircase pattern of \(L'\) allows us to solve easily by backwards substitution. In more detail:
  1. Equation 3 in \(L'\) tells us that \(z=2\text{.}\)
  2. Now substitute \(z=2\) into Equation 2 of \(L'\) and solve for \(y\) to get \(y=4\text{.}\)
  3. Finally, substitute \(y=4\) and \(z=2\) into Equation 1 of \(L'\)and solve for \(x\) to get \(x=4\text{.}\) We conclude that \((x,y,z)=(4,4,2)\) is a solution to \(L'\text{.}\)
  4. Furthermore, since at each step we had no choice in the matter (\(z\) must be equal to 2, \(y\) then must be equal to 4, etc.), we see that in fact \((4,4,2)\) is the only solution to \(L'\text{.}\)
Our method for solving a more complicated system, like \(L\) above, will be to transform it to a simpler system like \(L'\text{.}\)
Key point. In order for this method to work, we need to make sure that the transformed system has exactly the same solutions as the original system! To this end we will restrict our permissible transformations to the elementary operations define below.
Definition 1.1.11. Elementary operations on linear systems.
An elementary operation (or elementary row operation) is one of the three types of operations on linear systems described below.
Scalar multiplication
Multiply an equation by a nonzero number \(c\ne 0\text{:}\) i.e., replace the \(i\)-th equation \(e_i\) of the system with \(c\cdot e_i\) for \(c\ne 0\text{.}\) Note: \(c\cdot e_i\) is the result of multiplying the left and right sides of equation \(e_i\) by \(c\text{.}\)
Equation swap
Swap the \(i\)-th and \(j\)-th equations of the system, \(i\ne j\text{:}\) i.e., replace equation \(e_i\) of the system with \(e_j\text{,}\) and replace equation \(e_j\) with \(e_i\text{.}\)
Equation addition
Add a multiple of one equation to another: i.e., replace \(e_i\) with \(e_i+ce_j\) for some \(c\text{,}\) \(i\text{,}\) and \(j\text{.}\)
The act of transforming a system of equations using elementary operations is called reduction (or row reduction).
After reducing a linear system \(L\) using elementary operations, we are left with a new system \(L'\text{.}\) The systems \(L\) and \(L'\) will generally look very different from one another. Note, however, that the two systems will have the same number of equations, and the same number of variables. More importantly, the two systems will have identical sets of solutions; that is, the set of solutions of the new system \(L'\) is identical to the set of solutions to \(L\text{!}\)
To convince ourselves of this last assertion, it suffices to show that the application of any one of these elementary row operations produces a new system \(L'\) with exactly the same set of solutions as \(L\text{;}\) if this is so, then performing any finite sequence of elementary row operations must also preserve the set of solutions in this sense. Theorem 1.1.13 makes this argument official. First we introduce the notion of equivalent linear systems, mainly to spare ourselves from the mouthful that is “obtained by applying a finite sequence of elementary row operations”.
Definition 1.1.12. Row equivalent linear systems.
We say two systems of linear equations are row equivalent if the one can be obtained from the other by a finite sequence of elementary operations.
We prove by induction that if system \(L'\) is the result of applying \(n\) elementary operations to system \(L\text{,}\) \(n\geq 0\text{,}\) then \(L\) and \(L'\) have the same set of solutions.
Base case: \(n=0\).
In this case \(L=L'\) (we have applied no operations) and the statement is obvious.
Induction step.
Assume that applying a sequence of \(n\) elementary operations to a linear system preserves the set of solutions.
Suppose \(L'\) is the result of applying \(n+1\) elementary operations to the system \(L\text{.}\) Let \(L''\) be the result of applying the first \(n\) of these operations. By the induction hypothesis, systems \(L\) and \(L''\) have the same set of solutions.
Since the system \(L'\) is obtained from \(L''\) by applying exactly one elementary operation, it now suffices to show that applying a single row operation does not change the set of solutions to a linear system. This is the result of Exercise 1.1.3.7. (Note that a solution is provided to that exercise.)
We conclude that \(L\) and \(L'\) have the same set of solutions, as desired.

Example 1.1.14. Complete example.

Consider again the linear system
\begin{equation*} \begin{linsys}{3} 2x\amp +\amp 3y\amp +\amp -z\amp =\amp 18\\ x\amp +\amp 2y\amp -\amp 2z\amp =\amp 8\\ -\frac{1}{2}x\amp +\amp -\frac{1}{2}y\amp +\amp \frac{1}{2}z\amp =\amp -3 \end{linsys} \end{equation*}
We transform the system as as follows:
\begin{align*} \begin{linsys}{3} 2x\amp +\amp 3y\amp +\amp -z\amp =\amp 18\\ x\amp +\amp 2y\amp -\amp 2z\amp =\amp 8\\ -\frac{1}{2}x\amp +\amp -\frac{1}{2}y\amp +\amp \frac{1}{2}z\amp =\amp -3 \end{linsys} \amp \xrightarrow[\hspace{15pt}]{2e_3}\amp \begin{linsys}{3} 2x\amp +\amp 3y\amp +\amp -z\amp =\amp 18\\ x\amp +\amp 2y\amp -\amp 2z\amp =\amp 8\\ -x\amp +\amp -y\amp +\amp z\amp =\amp -6 \end{linsys}\\ \amp \\ \amp \xrightarrow[\hspace{10pt}]{e_1-e_2}\amp \begin{linsys}{3} x\amp +\amp y\amp +\amp z\amp =\amp 10\\ x\amp +\amp 2y\amp -\amp 2z\amp =\amp 8\\ -x\amp +\amp -y\amp +\amp z\amp =\amp -6 \end{linsys}\\ \amp \\ \amp \xrightarrow[\hspace{10pt}]{e_2-e_1}\amp \begin{linsys}{3} x\amp +\amp y\amp +\amp z\amp =\amp 10\\ \amp \amp y\amp -\amp 3z\amp =\amp -2\\ \amp \amp \amp \amp 2z\amp =\amp 4 \end{linsys}\\ \amp \\ \amp \xrightarrow[\hspace{10pt}]{e_3+e_1}\amp \begin{linsys}{3} x\amp +\amp y\amp +\amp z\amp =\amp 10\\ \amp \amp y\amp -\amp 3z\amp =\amp -2\\ \amp \amp \amp \amp 2z\amp =\amp 4 \end{linsys}\\ \amp \\ \amp \xrightarrow[]{\frac{1}{2}e_3}\amp \begin{linsys}{3} x\amp +\amp y\amp +\amp z\amp =\amp 10\\ \amp \amp y\amp -\amp 3z\amp =\amp -2\\ \amp \amp \amp \amp z\amp =\amp 2 \end{linsys} \end{align*}
Now put the logic together. Our original linear system \(L\) was transformed by a sequence of elementary row operations to a new system \(L'\text{:}\)
\begin{align*} \begin{linsys}{3} 2x\amp +\amp 3y\amp +\amp -z\amp =\amp 18\\ x\amp +\amp 2y\amp -\amp 2z\amp =\amp 8\\ -x\amp +\amp -y\amp +\amp z\amp =\amp -6 \end{linsys} \amp \xrightarrow[]{\text{ row op.'s } }\amp \begin{linsys}{3} x\amp +\amp y\amp +\amp z\amp =\amp 10\\ \amp \amp y\amp -\amp 3z\amp =\amp -2\\ \amp \amp \amp \amp z\amp =\amp 2 \end{linsys} \end{align*}
We saw already that the second system has exactly one solution, namely the triple \((x,y,z)=(4,4,2)\text{.}\)
Since transforming a system by row operations preserves solutions, the first and second systems have exactly the same solutions.
Thus \((x,y,z)=(4,4,2)\) is the only solution to the original system!

Remark 1.1.15. Notation.

As we will see later, keeping track of the exact sequence of row operations is important in some situations. The notation used in the last example is useful for this bookkeeping. Let’s explicate the notation somewhat.
The notation
\begin{equation*} L\xrightarrow{c\, e_i}L' \end{equation*}
indicates that system \(L'\) is obtained from \(L\) by replacing equation \(e_i\) with \(c\, e_i\text{.}\)
The notation
\begin{equation*} L\xrightarrow{e_i\leftrightarrow e_j}L' \end{equation*}
indicates that system \(L'\) is obtained from \(L\) by swapping rows \(e_i\) and \(e_j\text{.}\)
The notation
\begin{equation*} L\xrightarrow{e_i+c\,e_j}L' \end{equation*}
indicates that system \(L'\) is obtained from \(L\) by replacing equation \(e_i\) with \(e_i+c\,e_i\text{.}\)

Remark 1.1.16. Mandate.

You may be tempted to do multiple operations in a single step during this procedure. Resist this temptation for now until we have a better understanding of things.
Furthermore, if later on you do give in to this temptation, make sure that the two (or more row operations) you perform are independent of one another. For example, do not swap \(e_2\) with \(e_1\) and replace \(e_3\) with \(e_3-5e_2\) in the same step.

Exercises 1.1.3 Exercises

WeBWork Exercises

1.
Determine whether the following system has no solution, an infinite number of solutions or a unique solution.
  1. \(\displaystyle \left\lbrace\begin{array}{rllrrr} -8 x \amp - 7 y \amp = \amp 3 \cr 4 x \amp + 4 y \amp = \amp 8 \cr -16 x \amp - 13 y \amp = \amp 25 \end{array} \right.\)
  2. \(\displaystyle \left\lbrace\begin{array}{rllrrr} -8 x \amp - 7 y \amp = \amp 3 \cr 4 x \amp + 4 y \amp = \amp 8 \cr -16 x \amp - 13 y \amp = \amp 27 \end{array} \right.\)
  3. \(\displaystyle \left\lbrace\begin{array}{rllrrr} 25 x \amp - 15 y \amp = \amp 10 \cr -15 x \amp + 9 y \amp = \amp -6 \cr 30 x \amp - 18 y \amp = \amp 12 \end{array} \right.\)
2.
Suppose that the following
\(\left\lbrace\begin{array}{rrrrrr}\amp 4 \,x\amp +\amp 8 \, y\amp =\amp 4\\\amp 12 \,x\amp +\amp 24 \, y\amp =\amp k\\\amp 8 \,x\amp +\amp 16 \, y\amp =\amp 8\\ \end{array}\right.\)
is a consistent system. Then \(k=\)
Answer.
\(12\)
3.
Find the set of solutions for the linear system
\begin{equation*} \begin{array}{rcrcrcr} x_1 \amp + \amp 4 x_{2} \amp + \amp 6 x_{3} \amp = \amp -10 \\ \amp - \amp x_{2} \amp - \amp 4 x_{3} \amp = \amp -9 \\ \amp \amp \amp \amp 3 x_{3} \amp = \amp 9\end{array} \end{equation*}
Use s1, s2, etc. for the free variables if necessary.
\((x_1, x_2, x_3) = \bigg(\) , , \(\bigg)\)
Answer 1.
\(-16\)
Answer 2.
\(-3\)
Answer 3.
\(3\)
Solution.
SOLUTION: Equation 3 \(\Rightarrow x_3 = 3\text{.}\) Substitute into equation 2, \(-x_2 - 4(3) = -9 \Rightarrow x_2 = -3\text{.}\) Substitute into equation 1, \(x_1 + 4(-3) + 6(3) = -10 \Rightarrow x_1 = -16\text{.}\)
4.
Give a geometric description of the following systems of equations
  1. \(\displaystyle \left\lbrace\begin{array}{rrrrrr}\amp 7 \,x\amp +\amp \, y\amp =\amp 9\\\amp 4 \,x\amp +\amp 3 \, y\amp =\amp 2\\\amp -23 \, x\amp - \amp 13 \,y\amp =\amp -15\\ \end{array}\right.\)
  2. \(\displaystyle \left\lbrace\begin{array}{rrrrrr}\amp -8 \, x\amp +\amp 12 \, y\amp =\amp -4\\\amp -4 \, x\amp +\amp 6 \, y\amp =\amp -2\\\amp 14 \,x\amp - \amp 21 \,y\amp =\amp 7\\ \end{array}\right.\)
  3. \(\displaystyle \left\lbrace\begin{array}{rrrrrr}\amp 7 \,x\amp +\amp \, y\amp =\amp 9\\\amp 4 \,x\amp +\amp 3 \, y\amp =\amp 2\\\amp -23 \, x\amp - \amp 13 \,y\amp =\amp -17\\ \end{array}\right.\)
5.
Give a geometric description of the following system of equations
  1. \(\displaystyle \left\lbrace\begin{array}{rrrrrrr} 2x \amp + \amp 4y \amp - \amp 6z \amp = \amp -12 \cr -3x \amp - \amp 6y \amp + \amp 9z \amp = \amp 18 \end{array} \right.\)
  2. \(\displaystyle \left\lbrace\begin{array}{rrrrrrr} 2x \amp + \amp 4y \amp - \amp 6z \amp = \amp 12 \cr -x \amp + \amp 5y \amp - \amp 9z \amp = \amp 1 \end{array} \right.\)
  3. \(\displaystyle \left\lbrace\begin{array}{rrrrrrr} 2x \amp + \amp 4y \amp - \amp 6z \amp = \amp 12 \cr -3x \amp - \amp 6y \amp + \amp 9z \amp = \amp 16 \end{array}\right.\)

Written exercises

6. Geometry of linear systems.
In this exercise we will use the geometry of lines in \(\R^2\) and planes in \(\R^3\) to help analyze solutions to systems of equations in two and three unknowns, respectively. See Example 1.1.7 for a refresher on these concepts.
  1. Fix \(m\gt 1\) and consider a system of \(m\) linear equations in the two unknowns \(x\) and \(y\text{.}\) What does a solution \((x,y)\) to this system of linear equations correspond to geometrically?
  2. Use your interpretation in (a) to give a geometric argument that a system of \(m\) equations in two unknowns will have either (i) zero solutions, (ii) exactly one solution, or (iii) infinitely many solutions.
  3. Use your geometric interpretation to help produce explicit examples of systems in two variables satisfying these three different cases (i)-(iii).
  4. Now repeat (a)-(b) for a system of linear equations in three variables \(x,y, z\text{.}\)
Solution.
(a) Geometrically, each equation in the system represents a line \(\ell_i\colon a_ix+b_iy=c_i\text{.}\) A solution \((x,y)\) to the \(i\)-th equation corresponds to a point on \(\ell_i\text{.}\) Thus a solution \((x,y)\) to the system corresponds to a point lying on all of the lines: i.e., a point of intersection of the lines.
(b) First of all to prove the desired “or” statement it suffices to prove that if the number of solutions is greater than 1, then there are infinitely many solutions.
Now suppose there is more than one solution. Then there are at least two different solutions: \(P_1=(x_1,y_1)\) and \(P_2=(x_2,y_2)\text{.}\) Take any of the two lines \(\ell_i, \ell_j\text{.}\) By above the intersection of \(\ell_i\) and \(\ell_j\) contains \(P_1\) and \(P_2\text{.}\) But two distinct lines intersect in at most one point. It follows that \(\ell_i\) and \(\ell_j\) must be equal. Since \(\ell_i\) and \(\ell_j\) were arbitrary, it follows all of the lines \(\ell_i\) are in fact the same line \(\ell\text{.}\) But this means the common intersection of the lines is \(\ell\text{,}\) which has infinitely many points. It follows that the system has infinitely many solutions.
(c) We will get 0 solutions if the system includes two different parallel lines: e.g., \(\ell_1\colon x+y=5\) and \(\ell_2\colon x+y=1\text{.}\)
We will get exactly one solution when the slopes of each line in the system are distinct.
We will get infinitely many solutions when all equations in the system represent the same line. This happens when all equations are multiples of one another.
(d) Now each equation in our system defines a plane \(\mathcal{P}_i\colon a_ix+b_iy+c_iz=d_i\text{.}\) A solution \((x,y,z)\) to the system corresponds to a point \(P=(x,y,z)\) of intersection of the planes. We recall two facts from Euclidean geometry:
  1. Fact 1.
    Given two distinct points, there is a unique line containing both of them.
  2. Fact 2.
    Given any number of distinct planes, they either do not intersect, or intersect in a line.
We proceed as in part (b) above: that is show that if there are two distinct solutions to the system, then there are infinitely many solutions. First, for simplicity, we may assume that the equations \(\mathcal{P}_i\colon a_ix+b_iy+c_iz=d_i\) define distinct planes; if we have two equations defining the same plane, we can delete one of them and not change the set of solutions to the system.
Now suppose \(P=(x_1,y_1,z_1)\) and \(Q=(x_2,y_2,z_2)\) are two distinct solutions to the system. Let \(\ell\) be the unique line containing \(P\) and \(Q\) (Fact 1). I claim that \(\ell\) is precisely the set of solutions to the system. To see this, take any two equations in the system \(\mathcal{P}_i\colon a_ix+b_iy+c_iz=d_i\) and \(\mathcal{P}_j\colon a_jx+b_jy+c_iz=d_j\text{.}\) Since the two corresponding planes are distinct, and intersect in at least the points \(P\) and \(Q\text{,}\) they must intersect in a line (Fact 2); since this line contains \(P\) and \(Q\text{,}\) it must be the line \(\ell\) (Fact 1). Thus any two planes in the system intersect in the line \(\ell\text{.}\) From this it follows that: (a) a point satisfying the system must lie in \(\ell\text{;}\) and (b) all points on \(\ell\) satisfy the system (since we have shown that \(\ell\) lies in all the planes). It follows that \(\ell\) is precisely the set of solutions, and hence that there are infinitely many solutions.
7. Row operations preserve solutions.
We made the claim that each of our three row operations
  1. scalar multiplication (\(e_i\mapsto c\cdot e_i\) for \(c\ne 0\)),
  2. swap (\(e_i\leftrightarrow e_j\)),
  3. addition (\(e_i\mapsto e_i+c\cdot e_j\) for some \(c\))
do not change the set of solutions of a linear system. To prove this claim, let \(L\) be a general linear system
\begin{equation*} \numeqsys\text{.} \end{equation*}
Now consider each type of row operation separately, write down the new system \(L'\) you get by applying this row operation, and prove that an \(n\)-tuple \(s=(s_1,s_2,\dots ,s_n)\) is a solution to the original system \(L\) if and only if it is a solution to the new system \(L'\text{.}\)
Solution.
Let \(L\) be the original system with equations \(e_1,e_2,\dots ,e_m\text{.}\) For each specified row operation, we will call the resulting new system \(L'\) and its equations \(e'_1,e'_2,\dots , e'_m\text{.}\)
Row swap.
In this case systems \(L\) and \(L'\) have exactly the same equations, just written in a different order. Thus the \(n\)-tuple \(s\) satisfies \(L\) if and only if it satisfies each of the equations \(e_i\text{,}\) if and only if it satisfies each of the equations \(e'_i\text{,}\) since these are the same equations! It follows that \(s\) is a solution of \(L\) if and only if it is a solution to \(L'\text{.}\)
Scalar multiplication.
In this case \(e_j=e'_j\) for all \(j\ne i\text{,}\) and \(e'_i=c\cdot e_i\) for some \(c\ne 0\text{.}\) Since only the \(i\)-th equation has changed, it suffices to show that \(s\) is a solution to \(e_i\) if and only if \(s\) is a solution to \(c\cdot e_i\text{.}\) Let’s prove each direction of this if and only if separately.
If \(s\) satisfies \(e_1\text{,}\) then \(a_{i1}s_1+a_{i2}s_2+\cdots +a_{in}s_n=b_i\text{.}\) Multiplying both sides by \(c\) we see that
\begin{equation*} ca_{i1}s_1+ca_{i2}s_2+\cdots +ca_{in}s_n=cb_i, \end{equation*}
and hence that \(s\) is also a solution of \(c\,e_i=e_i'\text{.}\)
For the other direction, if \(s\) satisfies \(c e_i=e_i'\text{,}\) then
\begin{equation*} ca_{ii}s_1+ca_{i2}s_2+\cdots ca_{in}s_n=cb_i. \end{equation*}
Now, since \(c\ne 0\text{,}\) we can multiply both sides of this equation by \(1/c\) to see that
\begin{equation*} a_{i1}s_1+a_{i2}s_2+\cdots +a_{in}s_n=b_i \end{equation*}
and hence that \(s\) is a solution to \(e_i\text{.}\)
Row addition.
The only equation of \(L'\) that differs from \(L\) is
\begin{equation*} e_i'=e_i+ce_j\text{.} \end{equation*}
Writing this equation out in terms of coefficients gives us
\begin{equation*} e_i': a_{i1}x_1+a_{i2}x_2+\cdots +a_{in}x_n+c(a_{j1}x_1+a_{j2}x_2+\cdots +a_{jn}x_n)=b_i+cb_j\text{.} \end{equation*}
Now if \(s\) satisfies \(L\text{,}\) then it satisfies \(e_i\) and \(e_j\text{,}\) in which case evaluating the RHS of the \(e_i'\) above at \(s\) yields
\begin{align*} a_{i1}s_1+a_{i2}s_2+\cdots +a_{in}s_n+c(a_{j1}s_1+a_{j2}s_2+\cdots +a_{jn}s_n)\amp =b_i+cb_j \end{align*}
showing that \(s\) satisfies \(e_i'\text{.}\) Now suppose \(s=(s_1,s_2,\dots, s_n)\) satisfies \(L'\text{.}\) Since \(s\) satisfies \(e_j'=e_j\text{,}\) we have
\begin{equation} a_{j1}s_1+a_{j2}s_2+\cdots +a_{jn}s_n =b_j\tag{1.1.3} \end{equation}
Since \(s\) satisfies \(e_i'\text{,}\) we have
\begin{equation*} a_{i1}s_1+a_{i2}s_2+\cdots +a_{in}s_n+c(a_{j1}s_1+a_{j2}s_2+\cdots +a_{jn}s_n)=b_i+cb_j \end{equation*}
Substituting (1.1.3) into the equation above we get
\begin{equation*} a_{i1}s_1+a_{i2}s_2+\cdots +a_{in}s_n+c(b_j)=b_i+cb_j\text{,} \end{equation*}
and hence
\begin{equation*} a_{i1}s_1+a_{i2}s_2+\cdots +a_{in}s_n=b_i\text{.} \end{equation*}
This shows that \(s\) satisfies \(e_i\text{.}\) It follows that \(s\) satisfies \(L\text{.}\)
8. Nonlinear systems.
A nonlinear system of equations is a collection of equations, at least one of which is nonlinear. Our definition of a solution to a linear system generalizes easily to any system of equations.
  1. Consider the following nonlinear system in the unknowns \(x, y\text{:}\)
    \begin{equation*} \begin{linsys}{2} x^2\amp +\amp y^2\amp =\amp 1\\ x\amp +\amp y \amp =\amp \frac{1}{2} \end{linsys}\text{.} \end{equation*}
    1. Sketch the graphs of each of the two equations in the system on a common coordinate system.
    2. Describe geometrically what a solution to the system is in terms of your sketch. Explain your reasoning. How many solutions to the system are there, according to your sketch?
    3. Compute the set of all solutions to the system algebraically.
  2. Now consider a more general system
    \begin{equation} \begin{linsys}{2} x^2\amp +\amp y^2\amp =\amp 1\\ ax\amp +\amp by \amp =\amp c, \end{linsys}\tag{1.1.4} \end{equation}
    where \(a,b,c\in\R\) are fixed constants and at least one of \(a\) or \(b\) is nonzero.
    1. Explain geometrically what a solution to the system corresponds to in terms of the graphs of its two equations.
    2. Use your geometric interpretation in (i) to argue that the system (1.1.4) has either 0, 1, or 2 solutions. Give explicit examples of such systems corresponding to each of these three cases.
9. Not all arithmetic operations preserve solutions.
In this exercise we investigate how the operation of squaring both sides of an equation changes the set of solutions. Let
\begin{equation} F(x_1,x_2,\dots, x_n)=G(x_1,x_2,\dots, x_n)\tag{1.1.5} \end{equation}
represent a general equation (linear or nonlinear) in the unknowns \(x_1,x_2,\dots, x_n\text{,}\) let
\begin{equation} \left(F(x_1,x_2,\dots, x_n)\right)^2=\left(G(x_1,x_2,\dots, x_n)\right)^2\tag{1.1.6} \end{equation}
be the equation obtained by squaring both sides of the (1.1.5), let \(S_1\) be the set of solutions to (1.1.5), and let \(S_2\) be the set of solutions to (1.1.6).
  1. Show that \(S_1\subseteq S_2\text{.}\)
  2. Given an explicit example of an equation of the form (1.1.5) in two variables for which \(S_2\not\subseteq S_1\text{.}\)
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