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Theorem 7.2.11 The Pigeonhole Principle. Let \( f \) be a function from a finite set \( X \) into a finite set \( Y \) . If \( n \geq 1 \) and \( \left| X\right| > n\left| Y\right| \), then there exists an element of \( Y \) that is the image under \( f \) of at least \( n + 1 \) elements of \( X \) . | Proof. Assume no such element exists. For each \( y \in Y \), let \( {A}_{y} = \{ x \in X \mid \) \( f\left( x\right) = y\} \) . Then it must be that \( \left| {A}_{y}\right| \leq n \) . Furthermore, the set of nonempty \( {A}_{y} \) form a partition of \( X \) . Therefore,\n\n\[\n\left| X\right| = \mathop{\sum }\limit... | Yes |
A duplicate name is assured. Assume that a room contains four students with the first names John, James, and Mary. Prove that two students have the same first name. | We can visualize a mapping from the set of students to the set of first names; each student has a first name. The pigeonhole principle applies with \( n = 1 \), and we can conclude that at least two of the students have the same first name. | Yes |
Example 7.3.3 A basic example. Let \( f : \{ 1,2,3\} \rightarrow \{ a, b\} \) be defined by \( f\left( 1\right) = a, f\left( 2\right) = a \), and \( f\left( 3\right) = b \) . Let \( g : \{ a, b\} \rightarrow \{ 5,6,7\} \) be defined by \( g\left( a\right) = 5 \) and \( g\left( b\right) = 7 \) . Then \( g \circ f : \{ 1... | For example, \( \left( {g \circ f}\right) \left( 1\right) = \) \( g\left( {f\left( l\right) }\right) = g\left( a\right) = 5 \) . Note that \( f \circ g \) is not defined. Why? | Yes |
Theorem 7.3.4 Function composition is associative. If \( f : A \rightarrow B \) , \( g : B \rightarrow C \), and \( h : C \rightarrow D \), then \( h \circ \left( {g \circ f}\right) = \left( {h \circ g}\right) \circ f \) . | Proof. Note: In order to prove that two functions are equal, we must use the definition of equality of functions. Assuming that the functions have the same domain, they are equal if, for each domain element, the images of that element under the two functions are equal.\n\nWe wish to prove that \( \left( {h \circ \left(... | Yes |
The inverse of a function on \( \{ 1,2,3\} \) . Let \( A = \{ 1,2,3\} \) and let \( f \) be the function defined on \( A \) such that \( f\left( 1\right) = 2, f\left( 2\right) = 3 \), and \( f\left( 3\right) = 1 \) . | Then \( {f}^{-1} : A \rightarrow A \) is defined by \( {f}^{-1}\left( 1\right) = 3,{f}^{-1}\left( 2\right) = 1 \), and \( {f}^{-1}\left( 3\right) = 2 \) . | Yes |
Theorem 7.3.14 Bijections have inverses. Let \( f : A \rightarrow A.{f}^{-1} \) exists if and only if \( f \) is a bijection; i. e. \( f \) is one-to-one and onto. | Proof. \( \left( \Rightarrow \right) \) In this half of the proof, assume that \( {f}^{-1} \) exists and we must prove that \( f \) is one-to-one and onto. To do so, it is convenient for us to use the relation notation, where \( f\left( s\right) = t \) is equivalent to \( \left( {s, t}\right) \in f \) . To prove that \... | No |
Example 7.3.18 Another inverse. Let \( A = \{ 1,2,3\} \) and \( B = \{ a, b, c\} \) . Define \( f : A \rightarrow B \) by \( f\left( 1\right) = a, f\left( 2\right) = b \), and \( f\left( 3\right) = c \) . Then \( g : B \rightarrow A \) defined by \( g\left( a\right) = 1, g\left( b\right) = 2 \), and \( g\left( c\right)... | \[ \left. \begin{array}{l} \left( {g \circ f}\right) \left( 1\right) = 1 \\ \left( {g \circ f}\right) \left( 2\right) = 2 \\ \left( {g \circ f}\right) \left( 3\right) = 3 \end{array}\right\} \Rightarrow g \circ f = {i}_{A}\text{ and }\left. \begin{array}{l} \left( {f \circ g}\right) \left( a\right) = a \\ \left( {f \ci... | Yes |
Define the sequence of numbers \( B \) by\n\n\[ \n{B}_{0} = {100}\text{and} \n\]\n\n\[ \n{B}_{k} = {1.08}{B}_{k - 1}\text{ for }k \geq 1. \n\] | These rules stipulate that each number in the list is 1.08 times the previous number, with the starting number equal to 100 . For example\n\n\[ \n{B}_{3} = {1.08}{B}_{2} \n\]\n\n\[ \n= {1.08}\left( {{1.08}{B}_{1}}\right) \n\]\n\n\[ \n= {1.08}\left( {{1.08}\left( {{1.08}{B}_{0}}\right) }\right) \n\]\n\n\[ \n= {1.08}\lef... | Yes |
The Fibonacci sequence is the sequence \( F \) defined by\n\n\[ \n{F}_{0} = 1,{F}_{1} = 1\text{and} \n\]\n\n\[ \n{F}_{k} = {F}_{k - 2} + {F}_{k - 1}\text{ for }k \geq 2 \n\] | To determine, for example, the fourth item in the Fibonacci sequence we repeatedly apply the recursive rule for \( F \) until we are left with an expression involving \( {F}_{0} \) and \( {F}_{1} \) :\n\n\[ \n{F}_{4} = {F}_{2} + {F}_{3} \n\]\n\n\[ \n= \left( {{F}_{0} + {F}_{1}}\right) + \left( {{F}_{1} + {F}_{2}}\right... | Yes |
A formula for the sequence \( B \) in Example 8.1.7 is \( B = {100}{\left( {1.08}\right) }^{k} \) for \( k \geq 0 \). | If \( k = 0 \), then \( B = {100}{\left( {1.08}\right) }^{0} = {100} \), as defined. Now assume that for some \( k \geq 1 \), the formula for \( {B}_{k} \) is true.\n\n\[ \n{B}_{k + 1} = {1.08}{B}_{k}\text{by the recursive definition} \]\n\n\[ \n= {1.08}\left( {{100}{\left( {1.08}\right) }^{k}}\right) \text{by the indu... | Yes |
Example 8.3.8 First Order Homogeneous Recurrence Relations. \( D\left( k\right) - {2D}\left( {k - 1}\right) = 0 \) is a first-order homogeneous relation. | Since it can also be written as \( D\left( k\right) = {2D}\left( {k - 1}\right) \), it should be no surprise that it arose from an expression that involves powers of 2 . More generally, you would expect that the solution of \( L\left( k\right) - {aL}\left( {k - 1}\right) \) would involve \( {a}^{k} \) . Actually, the s... | Yes |
Consider the second-order homogeneous relation \( S\left( k\right) - {7S}\left( {k - 1}\right) + {12S}\left( {k - 2}\right) = 0 \) together with the initial conditions \( S\left( 0\right) = 4 \) and \( S\left( 1\right) = 4 \). | From our discussion above, we can predict that the solution to this relation involves terms of the form \( b{a}^{k} \), where \( b \) and \( a \) are nonzero constants that must be determined. If the solution were to equal this quantity exactly, then\n\n\[ S\left( k\right) = b{a}^{k} \]\n\n\[ S\left( {k - 1}\right) = b... | Yes |
Suppose that \( T \) is defined by \( T\left( k\right) = {7T}\left( {k - 1}\right) - {10T}\left( {k - 2}\right) \), with \( T\left( 0\right) = 4 \) and \( T\left( 1\right) = {17} \). | (a) Note that we have written the recurrence relation in \ | No |
Solve \( S\left( k\right) - {7S}\left( {k - 2}\right) + {6S}\left( {k - 3}\right) = 0 \), where \( S\left( 0\right) = 8, S\left( 1\right) = 6 \), and \( S\left( 2\right) = {22} \). | (a) The characteristic equation is \( {a}^{3} - {7a} + 6 = 0 \).\n\n(b) The only rational roots that we can attempt are \( \pm 1, \pm 2, \pm 3 \), and \( \pm 6 \). By checking these, we obtain the three roots 1, 2, and -3 .\n\n(c) The general solution is \( S\left( k\right) = {b}_{1}{1}^{k} + {b}_{2}{2}^{k} + {b}_{3}{\... | Yes |
Solve \( D\left( k\right) - {8D}\left( {k - 1}\right) + {16D}\left( {k - 2}\right) = 0 \), where \( D\left( 2\right) = {16} \) and \( D\left( 3\right) = {80} \). | (a) Characteristic equation: \( {a}^{2} - {8a} + {16} = 0 \).\n\n(b) \( {a}^{2} - {8a} + {16} = {\left( a - 4\right) }^{2} \). Therefore, there is a double characteristic root,\n\n(c) General solution: \( D\left( k\right) = \left( {{b}_{10} + {b}_{11}k}\right) {4}^{k} \).\n\n(d) \( \left\{ \begin{array}{l} D\left( 2\ri... | Yes |
Solve \( S\left( k\right) + {5S}\left( {k - 1}\right) = 9 \), with \( S\left( 0\right) = 6 \) . | (a) The associated homogeneous relation, \( S\left( k\right) + {5S}\left( {k - 1}\right) = 0 \) has the characteristic equation \( a + 5 = 0 \) ; therefore, \( a = - 5 \) . The homogeneous solution is \( {S}^{\left( h\right) }\left( k\right) = b{\left( -5\right) }^{k} \).\n\n(b) Since the right-hand side is a constant,... | Yes |
Consider \( T\left( k\right) - {7T}\left( {k - 1}\right) + {10T}\left( {k - 2}\right) = 6 + {8k} \) with \( T\left( 0\right) = 1 \) and \( T\left( 1\right) = 2 \) . | (a) From Example 8.3.13, we know that \( {T}^{\left( h\right) }\left( k\right) = {b}_{1}{2}^{k} + {b}_{2}{5}^{k} \) . Caution:Don’t apply the initial conditions to \( {T}^{\left( h\right) } \) until you add \( {T}^{\left( p\right) } \) !\n\n(b) Since the right-hand side is a linear polynomial, \( {T}^{\left( p\right) }... | Yes |
Suppose you open a savings account that pays an annual interest rate of \( 8\% \) . In addition, suppose you decide to deposit one dollar when you open the account, and you intend to double your deposit each year. Let \( B\left( k\right) \) be your balance after \( k \) years. \( B \) can be described by the relation \... | Returning to the original situation, (a) \( {B}^{\left( h\right) }\left( k\right) = {b}_{1}{\left( {1.08}\right) }^{k} \) (b) \( {B}^{\left( p\right) }\left( k\right) \) should be of the form \( d{2}^{k} \) . (c) \[ d{2}^{k} = {1.08d}{2}^{k - 1} + {2}^{k} \Rightarrow \left( {2d}\right) {2}^{k - 1} = {1.08d}{2}^{k - 1} ... | Yes |
Find the general solution to \( S\left( k\right) - 3S\left( k - 1\right) - 4S\left( k - 2\right) = 4^{k} \) | (a) The characteristic roots of the associated homogeneous relation are -1 and 4. Therefore, \( S^{\left( h\right) }\left( k\right) = b_{1}\left( -1\right)^{k} + b_{2}4^{k} \).\n\n(b) A function of the form \( d4^{k} \) will not be a particular solution of the nonhomogeneous relation since it solves the associated homo... | Yes |
Theorem 8.4.3 Fundamental Properties of Logarithms. Let \( a \) and \( b \) be positive real numbers, and \( r \) a real number. | \[ {\log }_{2}1 = 0 \] (8.4.4) \[ {\log }_{2}{ab} = {\log }_{2}a + {\log }_{2}b \] (8.4.5) \[ {\log }_{2}\frac{a}{b} = {\log }_{2}a - {\log }_{2}b \] (8.4.6) \[ {\log }_{2}{a}^{r} = r{\log }_{2}a \] (8.4.7) \[ {2}^{{\log }_{2}a} = a \] (8.4.8) | Yes |
Theorem 8.4.5 How logarithms with different bases are related. Let \( b > 0, b \neq 1 \) . Then for all \( a > 0,{\log }_{b}a = \frac{{\log }_{2}a}{{\log }_{2}b} \) . Therefore, if \( b > 1 \), base \( b \) logarithms can be computed from base 2 logarithms by dividing by the positive scaling factor \( {\log }_{2}b \) .... | Proof. By an analogue of (8.4.8), \( a = {b}^{{\log }_{b}a} \) . Therefore, if we take the base 2 logarithm of both sides of this equality we get:\n\n\[{\log }_{2}a = {\log }_{2}\left( {b}^{{\log }_{b}a}\right) \Rightarrow {\log }_{2}a = {\log }_{b}a \cdot {\log }_{2}b\]\n\nFinally, divide both sides of the last equati... | Yes |
(a) If \( {S}_{n} = {3}^{n}, n \geq 0 \), then\n\n\[ G\left( {S;z}\right) = 1 + {3z} + 9{z}^{2} + {27}{z}^{3} + \cdots \]\n\n\[ = \mathop{\sum }\limits_{{n = 0}}^{\infty }{3}^{n}{z}^{n} \]\n\n\[ = \mathop{\sum }\limits_{{n = 0}}^{\infty }{\left( 3z\right) }^{n} \] | We can get a closed form expression for \( G\left( {S;z}\right) \) by observing that \( G\left( {S;z}\right) - \) \( {3zG}\left( {S;z}\right) = 1 \) . Therefore, \( G\left( {S;z}\right) = \frac{1}{1 - {3z}} \). | Yes |
Example 8.5.8 Some operations on generating functions. If \( D\left( z\right) = \) \( \mathop{\sum }\limits_{{k = 0}}^{\infty }k{z}^{k} \) and \( H\left( z\right) = \mathop{\sum }\limits_{{k = 0}}^{\infty }{2}^{k}{z}^{k} \) then | \[ \left( {D + H}\right) \left( z\right) = \mathop{\sum }\limits_{{k = 0}}^{\infty }\left( {k + {2}^{k}}\right) {z}^{k} \] \[ \left( {2H}\right) \left( z\right) = \mathop{\sum }\limits_{{k = 0}}^{\infty }2 \cdot {2}^{k}{z}^{k} = \mathop{\sum }\limits_{{k = 0}}^{\infty }{2}^{k + 1}{z}^{k} \] \[ \left( {zD}\right) \left(... | Yes |
Theorem 8.5.9 Generating functions related to Pop and Push. If \( p > 1 \)\n\n(a) \( G\left( {S \uparrow p;z}\right) = \left( {G\left( {S;z}\right) - \mathop{\sum }\limits_{{k = 0}}^{{p - 1}}S\left( k\right) {z}^{k}}\right) /{z}^{k} \) | Proof. We prove (a) by induction and leave the proof of (b) to the reader.\n\nBasis:\n\n\[ G\left( {S \uparrow ;z}\right) = \mathop{\sum }\limits_{{k = 0}}^{\infty }S\left( {k + 1}\right) {z}^{k} \]\n\n\[ = \mathop{\sum }\limits_{{k = 1}}^{\infty }S\left( k\right) {z}^{k - 1} \]\n\n\[ = \left( {\mathop{\sum }\limits_{{... | No |
Solve \( S\left( k\right) + {3S}\left( {k - 1}\right) - \) \( {4S}\left( {k - 2}\right) = 0, k \geq 2 \), with \( S\left( 0\right) = 3 \) and \( S\left( 1\right) = - 2 \). | (1) Translate to an equation about generating functions. First, we change the index of the recurrence relation by substituting \( n + 2 \) for \( k \) . The result is \( S\left( {n + 2}\right) + {3S}\left( {n + 1}\right) - {4S}\left( n\right) = 0, n \geq 0 \) . Now, if \( V\left( n\right) = \) \( S\left( {n + 2}\right)... | Yes |
Suppose that you roll a die two times and add up the numbers on the top face for each roll. Since the faces on the die represent the integers 1 through 6, the sum must be between 2 and 12. How many ways can any one of these sums be obtained? | Obviously, 2 can be obtained only one way, with two 1's. There are two sequences that yield a sum of 3: 1-2 and 2-1. To obtain all of the frequencies with which the numbers 2 through 12 can be obtained, we set up the situation as follows. For \( j = 1,2;{P}_{j} \) is the rolling of the die for the \( {j}^{\text{th }} \... | Yes |
Example 8.5.15 Distribution of a Committee. Suppose that an organization is divided into three geographic sections, A, B, and C. Suppose that an executive committee of 11 members must be selected so that no more than 5 members from any one section are on the committee and that Sections A, B, and C must have minimums of... | Let \( {P}_{A} \) be the action of deciding how many members (not who) from Section A will serve on the committee. \( {X}_{A} = \{ 3,4,5\} \) and \( {Q}_{A}\left( k\right) = k \) . The frequency function, \( {F}_{A} \), is defined by \( {F}_{A}\left( k\right) = 1 \) if \( k \in {X}_{k} \), with \( {F}_{A}\left( k\right... | Yes |
Example 9.1.9 An Undirected Graph. A network of computers can be described easily using a graph. Figure 9.1.10 describes a network of five computers, \( a, b, c, d \), and \( e \) . An edge between any two vertices indicates that direct two-way communication is possible between the two computers. Note that the edges of... | This undirected graph, in set terms, is \( V = \{ a, b, c, d, e\} \) and \( E = \{ \{ a, b\} ,\{ a, d\} ,\{ b, c\} ,\{ b, d\} ,\{ c, e\} ,\{ b, e\} \} | Yes |
Suppose that a path between two vertices has an edge list \( \left( {{e}_{1},{e}_{2},\ldots ,{e}_{n}}\right) \) . A subpath of this graph is any portion of the path described by one or more consecutive edges in the edge list. For example, \( \left( {3,\mathrm{{No}},4}\right) \) is a subpath of \( \left( {1,2,3,\mathrm{... | A path or circuit is simple if it contains no proper subpath that is a circuit. This is the same as saying that a path or circuit is simple if it does not visit any vertex more than once except for the common initial and terminal vertex in the circuit. In the problem-solving method described in Figure 9.1.14, the path ... | No |
If you ignore the duplicate names of vertices in the four graphs of Figure 9.1.18, and consider the whole figure as one large graph, then there are four connected components in that graph. | It's as simple as that! It's harder to describe precisely than to understand the concept. | No |
A Very Small Example. We consider the representation of the following graph: | The adjacency matrix that represents the graph would be\n\n\[ \nG = \left( \begin{array}{llll} 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{array}\right) \]\n\nThe same graph could be represented with the edge dictionary\n\n\[ \n\{ 1 : \left\lbrack {2,4}\right\rbrack ,2 : \left\lbrack {3,4}\rig... | Yes |
Theorem 9.3.2 Maximal Path Theorem. If a graph has \( n \) vertices and vertex \( u \) is connected to vertex \( w \), then there exists a path from \( u \) to \( w \) of length no more than \( n \) . | Proof. (Indirect): Suppose \( u \) is connected to \( w \), but the shortest path from \( u \) to \( w \) has length \( m \), where \( m > n \) . A vertex list for a path of length \( m \) will have \( m + 1 \) vertices. This path can be represented as \( \left( {{v}_{0},{v}_{1},\ldots ,{v}_{m}}\right) \), where \( {v}... | Yes |
Example 9.3.10 A simple example. Consider the graph below. The existence of a path from vertex 2 to vertex 3 is not difficult to determine by examination. After a few seconds, you should be able to find two paths of length four. Algorithm 9.3.8 will produce one of them. | Suppose that the edges from each vertex are sorted in ascending order by terminal vertex. For example, the edges from vertex 3 would be in the order \( \left( {3,1}\right) ,\left( {3,4}\right) ,\left( {3,5}\right) \) . In addition, assume that in the body of Step 4 of the algorithm, the elements of \( {D}_{r} \) are us... | Yes |
If we compute all distances between vertices, we can summarize the results in a distance matrix, where the entry in row \( i \), column \( j \) is the distance from vertex \( i \) to vertex \( j \) . For the graph in Example 9.3.12, that matrix is\n\n\[ \left( \begin{matrix} 0 & 2 & 2 & 2 & 3 & 1 & 1 & 3 & 3 & 1 & 2 & ... | If we scan the matrix, we can see that the maximum distance is the distance between vertices 3 and 5 , which is 5 and is the diameter of the graph. If we focus on individual rows and identify the maximum values, which are the eccentricities, their minimum is 3 , which the graph's radius. This eccentricity value is atta... | Yes |
Theorem 9.4.3 Euler's Theorem: Koenigsberg Case. No walking tour of Koenigsberg can be designed so that each bridge is used exactly once. | Proof. The map of Koenigsberg can be represented as an undirected multigraph, as in Figure 9.4.2. The four land masses are the vertices and each edge represents a bridge.\n\nThe desired tour is then a path that uses each edge once and only once. Since the path can start and end at two different vertices, there are two ... | Yes |
A common problem encountered in engineering is that of analog-to-digital (a-d) conversion, where the reading on a dial, for example, must be converted to a numerical value. In order for this conversion to be done reliably and quickly, one must solve an interesting problem in graph theory. Suppose a dial can be turned i... | All digital computers represent numbers in binary form, as a sequence of 0 's and 1's called bits, short for binary digits. The binary representations of numbers 0 through 7 are:\n\n\[ 0 = {000}_{\text{two }} = 0 \cdot 4 + 0 \cdot 2 + 0 \cdot 1 \]\n\n\[ 1 = {001}_{\text{two }} = 0 \cdot 4 + 0 \cdot 2 + 1 \cdot 1 \]\n\n... | Yes |
The problem of a Boston salesman. The Traveling Salesman Problem gets its name from the situation of a salesman who wants to minimize the number of miles that he travels in visiting his customers. For example, if a salesman from Boston must visit the other capital cities of New England, then the problem is to find a ci... | The search for an efficient algorithm that solves the Traveling Salesman has occupied researchers for years. If the graph in question is complete, there are \( \left( {n - 1}\right) \) ! different circuits. As \( n \) gets large, it is impossible to check every possible circuit. The most efficient algorithms for solvin... | No |
Example 9.5.8 The One-way Street. A salesman must make stops at vertices A, B, and C, which are all on the same one-way street. The graph in Figure 9.5.9 is weighted by the function \( w\left( {i, j}\right) \) equal to the time it takes to drive from vertex \( i \) to vertex \( j \) . | Note that if \( j \) is down the one-way street from \( i \), then \( w\left( {i, j}\right) < w\left( {j, i}\right) \) . The values of \( {C}_{opt} \), and \( {C}_{cn} \) are 20 and 32, respectively. Verify that \( {C}_{cn} \) is 32 by using the closest neighbor algorithm. The value of \( \frac{{C}_{cn}}{{C}_{opt}} = {... | No |
Example 9.5.12 The Unit Square Problem. Suppose a robot is programmed to weld joints on square metal plates. Each plate must be welded at prescribed points on the square. To minimize the time it takes to complete the job, the total distance that a robot's arm moves should be minimized. Let \( d\left( {P, Q}\right) \) b... | Heuristic 9.5.13 The Strip Algorithm. Given n points in the unit square: Phase 1:\n\n(1) Divide the square into \( \left\lceil \sqrt{n/2}\right\rceil \) vertical strips, as in Figure 9.5.14. Let \( d \) be the width of each strip. If a point lies on a boundary between two strips, consider it part of the left-hand strip... | Yes |
Theorem 9.5.20 Flow out of Source equals Flow in Sink. If \( f \) is a flow, then \( \;\mathop{\sum }\limits_{{\left( {\text{source }, v}\right) \in E}}f\left( {\text{source }, v}\right) = \mathop{\sum }\limits_{{\left( {v,\text{ sink }}\right) \in E}}f\left( {v,\text{ sink }}\right) \) | Proof. Subtract the right-hand side of (9.5.1) from the left-hand side. The result is:\n\n\[ \text{Flow into}v - \text{Flow out of}v = 0 \]\n\nNow sum up these differences for each vertex in \( {V}^{\prime } = V - \{ \) source, sink \( \} \) . The result is\n\n\[ \mathop{\sum }\limits_{{v \in {V}^{\prime }}}\left( {\ma... | Yes |
Example 9.5.23 Augmenting City Water Flow. For \( {f}_{1} \) in Figure 9.5.18, a flow augmenting path would be \( \left( {{e}_{2},{e}_{3},{e}_{4}}\right) \) since \( w\left( {e}_{2}\right) - {f}_{1}\left( {e}_{2}\right) = {15}, w\left( {e}_{3}\right) - \) \( {f}_{1}\left( {e}_{3}\right) = 5 \), and \( w\left( {e}_{4}\r... | These positive differences represent unused capacities, and the smallest value represents the amount of flow that can be added to each edge in the path. Note that by adding 5 to each edge in our path, we obtain \( {f}_{2} \), which is maximal. If an edge with a positive flow is used in its reverse direction, it is cont... | Yes |
Example 9.5.27 A flow augmenting path going against the flow. Consider the network in Figure 9.5.28, where the current flow, \( f \), is indicated by a labeling of the edges. | The path (Source, \( {v}_{2},{v}_{1},{v}_{3},\operatorname{Sink} \) ) is a flow augmenting path that allows us to increase the flow by one unit. Note that \( \left( {{v}_{1},{v}_{3}}\right) \) is used in the reverse direction, which is allowed because \( f\left( {{v}_{1},{v}_{3}}\right) > 0 \) . The value of the new fl... | Yes |
Theorem 9.6.8 Euler’s Formula. If \( G = \left( {V, E}\right) \) is a connected planar graph with \( r \) regions, \( v \) vertices, and \( e \) edges, then\n\n\[ v + r - e = 2 \] | Proof. We prove Euler’s Formula by Induction on \( e \), for \( e \geq 0 \). \n\nBasis: If \( e = 0 \), then \( G \) must be a graph with one vertex, \( v = 1 \) ; and there is one infinite region, \( r = 1 \) . Therefore, \( v + r - e = 1 + 1 - 0 = 2 \), and the basis is true.\n\nInduction: Suppose that \( G \) has \(... | No |
Theorem 9.6.10 A Bound on Edges of a Planar Graph. If \( G = \left( {V, E}\right) \) is a connected planar graph with \( v \) vertices, \( v \geq 3 \), and \( e \) edges, then\n\n\[ e \leq {3v} - 6 \] | Proof. (Outline of a Proof)\n\n(a) Let \( r \) be the number of regions in \( G \) . For each region, count the number of edges that comprise its border. The sum of these counts must be at least \( {3r} \) . Recall that we are working with simple graphs here, so a region made by two edges connecting the same two vertic... | No |
Theorem 9.6.12 A Vertex of Degree Five. If \( G \) is a connected planar graph, then it has a vertex with degree 5 or less. | Proof. (by contradiction): We can assume that \( G \) has at least seven vertices, for otherwise the degree of any vertex is at most 5 . Suppose that \( G \) is a connected planar graph and each vertex has a degree of 6 or more. Then, since each edge contributes to the degree of two vertices, \( e \geq \frac{6v}{2} = {... | Yes |
Theorem 9.6.16 The Five-Color Theorem. If \( G \) is a planar graph, then \( \chi \left( G\right) \leq 5 \) . | Proof. The number 5 is not a sharp upper bound for \( \chi \left( G\right) \) because of the Four-Color Theorem.\n\nThis is a proof by Induction on the Number of Vertices in the Graph.\n\nBasis: Clearly, a graph with one vertex has a chromatic number of 1 .\n\nInduction: Assume that all planar graphs with \( n - 1 \) v... | Yes |
Theorem 9.6.20 No Odd Circuits in a Bipartite Graph. An undirected graph is bipartite if and only if it has no circuit of odd length. | Proof. \( \left( \Rightarrow \right) \) Let \( G = \left( {V, E}\right) \) be a bipartite graph that is partitioned into two sets, \( \mathrm{R}\left( \mathrm{{ed}}\right) \) and \( \mathrm{B}\left( \mathrm{{lue}}\right) \) that define a 2-coloring. Consider any circuit in \( V \) . If we specify a direction in the cir... | Yes |
Lemma 10.1.10 Let \( G = \left( {V, E}\right) \) be an undirected graph with no self-loops, and let \( {v}_{a},{v}_{b} \in V \) . If two different simple paths exist between \( {v}_{a} \) and \( {v}_{b} \), then there exists a cycle in \( G \) . | Proof. Let \( {p}_{1} = \left( {{e}_{1},{e}_{2},\ldots ,{e}_{m}}\right) \) and \( {p}_{2} = \left( {{f}_{1},{f}_{2},\ldots ,{f}_{n}}\right) \) be two different simple paths from \( {v}_{a} \) to \( {v}_{b} \) . The first step we will take is to delete from \( {p}_{1} \) and \( {p}_{2} \) the initial edges that are iden... | Yes |
Theorem 10.1.11 Equivalent Conditions for a Graph to be a Tree. Let \( G = \left( {V, E}\right) \) be an undirected graph with no self-loops and \( \left| V\right| = n \) . The following are all equivalent:\n\n(1) \( G \) is a tree.\n\n(2) For each pair of distinct vertices in \( V \), there exists a unique simple path... | Proof. Proof Strategy. Most of this theorem can be proven by proving the following chain of implications: \( \left( 1\right) \Rightarrow \left( 2\right) ,\left( 2\right) \Rightarrow \left( 3\right) ,\left( 3\right) \Rightarrow \left( 4\right) \), and \( \left( 4\right) \Rightarrow \left( 1\right) \) . Once these implic... | No |
Theorem 10.2.6 Let \( G = \left( {V, E, w}\right) \) be a weighted connected undirected graph. Let \( V \) be partitioned into two sets \( L \) and \( R \) . If \( {e}^{ * } \) is a bridge of least weight between \( L \) and \( R \), then there exists a minimal spanning tree for \( G \) that includes \( {e}^{ * } . | Proof. Suppose that no minimal spanning tree including \( {e}^{ * } \) exists. Let \( T = \) \( \left( {V,{E}^{\prime }}\right) \) be a minimal spanning tree. If we add \( {e}^{ * } \) to \( T \), a cycle is created, and this cycle must contain another bridge, \( e \), between \( L \) and \( R \) . Since \( w\left( {e}... | Yes |
Example 10.2.11 A Small Example. Consider the graph in Figure 10.2.12. If we apply Prim’s Algorithm starting at \( a \), we obtain the following edge list in the order given: \( \{ a, f\} ,\{ f, e\} ,\{ e, c\} ,\{ c, d\} ,\{ f, b\} ,\{ b, g\} \) . The total of the weights of these edges is 20 . | The method that we have used (in Step 2.1) to select a bridge when more than one minimally weighted bridge exists is to order all bridges alphabetically by the vertex in \( L \) and then, if further ties exist, by the vertex in \( R \) . The first vertex in that order is selected in Step 2.1 of the algorithm. | No |
The Case for Complete Graphs. The Minimum Diameter Spanning Tree Problem is trivial to solve in a \( {K}_{n} \) . Select any vertex \( {v}_{0} \) and construct the spanning tree whose edge set is the set of edges that connect \( {v}_{0} \) to the other vertices in the \( {K}_{n} \) . | Figure 10.2.15 illustrates a solution for \( n = 5 \) . | No |
Binary Tree Sort. Given a collection of integers (or other objects than can be ordered), one technique for sorting is a binary tree sort. If the integers are \( {a}_{1},{a}_{2},\ldots ,{a}_{n}, n \geq 1 \), we first execute the following algorithm that creates a binary tree: | Algorithm 10.4.8 Binary Sort Tree Creation.\n\n(1) Insert \( {a}_{1} \) into the root of the tree.\n\n(2) For \( k \mathrel{\text{:=}} 2 \) to \( n// \) insert \( {a}_{k} \) into the tree\n\n(a) \( r = {a}_{1} \)\n\n(b) inserted \( = \) false\n\n(c) while not(inserted):\n\nif \( {a}_{k} < r \) :\n\nif \( r \) has a lef... | Yes |
Theorem 11.2.1 A Monoid Theorem. If \( a, b \) are elements of \( M \) and \( a * b = b * a \), then \( \left( {a * b}\right) * \left( {a * b}\right) = \left( {a * a}\right) * \left( {b * b}\right) \). | Proof.\n\n\[ \left( {a * b}\right) * \left( {a * b}\right) = a * \left( {b * \left( {a * b}\right) }\right) \;\text{ Why? } \]\n\n\[ = a * \left( {\left( {b * a}\right) * b}\right) \;\text{ Why? } \]\n\n\[ = a * \left( {\left( {a * b}\right) * b}\right) \;\text{ Why? } \]\n\n\[ = a * \left( {a * \left( {b * b}\right) }... | No |
Consider the set of \( 2 \times 2 \) real matrices, \( {M}_{2 \times 2}\left( \mathbb{R}\right) \), with the operation of matrix multiplication. In this context, Theorem 11.2.1 can be interpreted as saying that if \( {AB} = {BA} \), then \( {\left( AB\right) }^{2} = {A}^{2}{B}^{2} \). | One pair of matrices that this theorem applies to is \( \left( \begin{array}{ll} 2 & 1 \\ 1 & 2 \end{array}\right) \) and \( \left( \begin{matrix} 3 & - 4 \\ - 4 & 3 \end{matrix}\right) \). | No |
Theorem 11.3.2 Identities are Unique - Rephrased. If \( G = \left\lbrack {G; * }\right\rbrack \) is a group and \( e \) is an identity of \( G \), then no other element of \( G \) is an identity of \( G \) . | Proof. (Indirect): Suppose that \( f \in G, f \neq e \), and \( f \) is an identity of \( G \) . We will show that \( f = e \), which is a contradiction, completing the proof.\n\n\[ f = f * e\;\text{Since}e\text{is an identity} \]\n\n\[ = e\;\text{Since}f\text{is an identity} \] | Yes |
Theorem 11.3.6 Inverse of Inverse Theorem (Rephrased). If a has inverse \( b \) and \( b \) has inverse \( c \), then \( a = c \) . | Proof.\n\n\[ a = a * e\;e\text{ is the identity of }G \]\n\n\[ = a * \left( {b * c}\right) \;\text{because}c\text{is the inverse of}b \]\n\n\[ = \left( {a * b}\right) * c\;\text{ why? } \]\n\n\[ = e * c\;\text{why?} \]\n\n\[ = c\;\text{by the identity property} \] | No |
Theorem 11.3.7 Inverse of a Product. If \( a \) and \( b \) are elements of group \( G \), then \( {\left( a * b\right) }^{-1} = {b}^{-1} * {a}^{-1} \) . | Proof. Let \( x = {b}^{-1} * {a}^{-1} \) . We will prove that \( x \) inverts \( a * b \) . Since we know\n\nthat the inverse is unique, we will have proved the theorem.\n\n\[ \left( {a * b}\right) * x = \left( {a * b}\right) * \left( {{b}^{-1} * {a}^{-1}}\right) \]\n\n\[ = a * \left( {b * \left( {{b}^{-1} * {a}^{-1}}\... | Yes |
Theorem 11.3.8 Cancellation Laws. If \( a, b \), and \( c \) are elements of group \( G \), then\n\n\[ \n\text{left cancellation:}\;\left( {a * b = a * c}\right) \Rightarrow b = c \n\]\n\n\[ \n\text{right cancellation:}\;\left( {b * a = c * a}\right) \Rightarrow b = c \n\] | Proof. We will prove the left cancellation law. The right law can be proved in exactly the same way. Starting with \( a * b = a * c \), we can operate on both \( a * b \) and \( a * c \) on the left with \( {a}^{-1} \) :\n\n\[ \n{a}^{-1} * \left( {a * b}\right) = {a}^{-1} * \left( {a * c}\right) \n\]\n\nApplying the as... | Yes |
Theorem 11.3.9 Linear Equations in a Group. If \( G \) is a group and \( a, b \in G \), the equation \( a * x = b \) has a unique solution, \( x = {a}^{-1} * b \) . In addition, the equation \( x * a = b \) has a unique solution, \( x = b * {a}^{-1} \) . | Proof. We prove the theorem only for \( a * x = b \), since the second statement is proven identically.\n\n\[ a * x = b = e * b \]\n\n\[ = \left( {a * {a}^{-1}}\right) * b \]\n\n\[ = a * \left( {{a}^{-1} * b}\right) \]\n\nBy the cancellation law, we can conclude that \( x = {a}^{-1} * b \) .\n\nIf \( c \) and \( d \) a... | Yes |
In the group of positive real numbers with multiplication, compute \( {5}^{3} \) and \( {5}^{-3} \). | \[ {5}^{3} = {5}^{2} \cdot 5 = \left( {{5}^{1} \cdot 5}\right) \cdot 5 = \left( {\left( {{5}^{0} \cdot 5}\right) \cdot 5}\right) \cdot 5 = \left( {\left( {1 \cdot 5}\right) \cdot 5}\right) \cdot 5 = 5 \cdot 5 \cdot 5 = {125} \] and \[ {5}^{-3} = {\left( {125}\right) }^{-1} = \frac{1}{125} \] | Yes |
Lemma 11.3.13 Let \( G \) be a group. If \( b \in G \) and \( n \geq 0 \), then \( {b}^{n + 1} = b * {b}^{n} \) , and hence \( b * {b}^{n} = {b}^{n} * b \) . | Proof. (By induction): If \( n = 0 \) ,\n\n\( {b}^{1} = {b}^{0} * b \) by the definition of exponentiation\n\n\( = e * b \) by the basis for exponentiation\n\n\( = b * e \) by the identity property\n\n\( = b * {b}^{0} \) by the basis for exponentiation\n\nNow assume the formula of the lemma is true for some \( n \geq 0... | Yes |
Theorem 11.3.14 Properties of Exponentiation. If a is an element of a group \( G \), and \( m \) and \( n \) are integers,\n\n(1) \( {a}^{-n} = {\left( {a}^{-1}\right) }^{n} \) and hence \( {\left( {a}^{n}\right) }^{-1} = {\left( {a}^{-1}\right) }^{n} \)\n\n(2) \( {a}^{n + m} = {a}^{n} * {a}^{m} \)\n\n(3) \( {\left( {a... | Proof. We will leave the proofs of these properties to the reader. All three parts can be done by induction. For example the proof of the second part would start by defining the proposition \( p\left( m\right), m \geq 0 \), to be \( {a}^{n + m} = {a}^{n} * {a}^{m} \) for all \( n \) . The basis is \( p\left( 0\right) :... | No |
Theorem 11.3.15 If \( G \) is a finite group, \( \left| G\right| = n \), and \( a \) is an element of \( G \) , then there exists a positive integer \( m \) such that \( {a}^{m} = e \) and \( m \leq n \) . | Proof. Consider the list \( a,{a}^{2},\ldots ,{a}^{n + 1} \) . Since there are \( n + 1 \) elements of \( G \) in this list, there must be some duplication. Suppose that \( {a}^{p} = {a}^{q} \), with \( p < q \) . Let \( m = q - p \) . Then\n\n\[ \n{a}^{m} = {a}^{q - p} \n\]\n\n\[ \n= {a}^{q} * {a}^{-p} \n\]\n\n\[ \n= ... | Yes |
Theorem 11.4.9 If \( a \) and \( b \) are positive integers, the smallest positive value of \( {ax} + {by} \) is the greatest common divisor of \( a \) and \( b,\gcd \left( {a, b}\right) \) . | Proof. If \( g = \gcd \left( {a, b}\right) \), since \( g\left| {a\text{and}g}\right| b \), we know that \( g \mid \left( {{ax} + {by}}\right) \) for any integers \( x \) and \( y \), so \( {ax} + {by} \) can’t be less than \( g \) . To show that \( g \) is exactly the least positive value, we show that \( g \) can be ... | Yes |
Theorem 11.4.15 Additive Inverses in \( {\mathbb{Z}}_{n} \) . If \( a \in {\mathbb{Z}}_{n}, a \neq 0 \), then the additive inverse of \( a \) is \( n - a \) . | Proof. \( a + \left( {n - a}\right) = n \equiv 0\left( {\;\operatorname{mod}\;n}\right) \), since \( n = n \cdot 1 + 0 \) . Therefore, \( a{ + }_{n}\left( {n - a}\right) = \) 0. | Yes |
Theorem 11.5.3 Subgroup Conditions. To determine whether \( H \), a subset of group \( \left\lbrack {G; * }\right\rbrack \), is a subgroup, it is sufficient to prove:\n\n(a) \( H \) is closed under \( * \) ; that is, \( a, b \in H \Rightarrow a * b \in H \) ;\n\n(b) \( H \) contains the identity element for \( * \) ; a... | Proof. Our proof consists of verifying that if the three properties above are true, then all the axioms of a group are true for \( \left\lbrack {H; * }\right\rbrack \) . By Condition (a), \( * \) can be considered an operation on \( H \) . The associative, identity, and inverse properties are the axioms that are needed... | Yes |
We can verify that \( 2\mathbb{Z} \leq \mathbb{Z} \), as stated in Example 11.5.2. | Whenever you want to discuss a subset, you must find some convenient way of describing its elements. An element of \( 2\mathbb{Z} \) can be described as 2 times an integer; that is, \( a \in 2\mathbb{Z} \) is equivalent to \( {\left( \exists k\right) }_{\mathbb{Z}}\left( {a = {2k}}\right) \) . Now we can verify that th... | Yes |
Theorem 11.5.5 Condition for a Subgroup of Finite Group. Given that \( \left\lbrack {G; * }\right\rbrack \) is a finite group and \( H \) is a nonempty subset of \( G \), if \( H \) is closed under \( * \), then \( H \) is a subgroup of \( G \) . | Proof. In this proof, we demonstrate that Conditions (b) and (c) of Theorem 11.5.3 follow from the closure of \( H \) under \( * \), which is condition (a) of the theorem. First, select any element of \( H \) ; call it \( \beta \) . The powers of \( \beta : {\beta }^{1},{\beta }^{2} \) , \( {\beta }^{3},\ldots \) are a... | Yes |
Example 11.6.2 A Direct Product of Monoids. Consider the monoids \( \mathbb{N} \) (the set of natural numbers with addition) and \( {B}^{ * } \) (the set of finite strings of \( 0 \) ’s and 1’s with concatenation). The direct product of \( \mathbb{N} \) with \( {B}^{ * } \) is a monoid. We illustrate its operation, whi... | \[ \left( {4,{001}}\right) * \left( {3,{11}}\right) = \left( {4 + 3,{001} + {11}}\right) = \left( {7,{00111}}\right) \] \[ \left( {0,{11010}}\right) * \left( {3,{01}}\right) = \left( {3,{1101001}}\right) \] \[ \left( {0,\lambda }\right) * \left( {{129},{00011}}\right) = \left( {0 + {129},\lambda + {00011}}\right) = \le... | Yes |
Theorem 11.6.5 The Direct Product of Groups is a Group. The direct product of two or more groups is a group; that is, the algebraic properties of a system obtained by taking the direct product of two or more groups includes the group axioms. | Proof. We will only present the proof of this theorem for the direct product of two groups. Some slight revisions can be made to produce a proof for any number of factors.\n\nStating that the direct product of two groups is a group is a short way of saying that if \( \\left\\lbrack {{G}_{1};{ * }_{1}}\\right\\rbrack \)... | Yes |
Example 11.7.2 How to Do Greek Arithmetic. Imagine that you are a six-year-old child who has been reared in an English-speaking family, has moved to Greece, and has been enrolled in a Greek school. Suppose that your new teacher asks the class to do the following addition problem that has been written out in Greek.\n\n\... | The natural thing for you to do is to take out your Greek-English/English-Greek dictionary and translate the Greek words to English, as outlined in Figure 11.7.3 After you've solved the problem, you can consult the same dictionary to find the proper Greek word that the teacher wants. Although this is not the recommende... | Yes |
Software Implementation of Sets. In this example, we will describe how set variables can be implemented on a computer. We will describe the two systems first and then describe the isomorphism between them. | System 1: The power set of \( \{ 1,2,3,4,5\} \) with the operation union, \( \cup \) . For simplicity, we will only discuss union. However, the other operations are implemented in a similar way.\n\nSystem 2: Strings of five bits of computer memory with an OR gate. Individual bit values are either zero or one, so the el... | Yes |
Example 11.7.7 Multiplying without doing multiplication. This isomorphism is between \( \left\lbrack {{\mathbb{R}}^{ + }; \cdot }\right\rbrack \) and \( \left\lbrack {\mathbb{R}; + }\right\rbrack \) . Until the 1970s, when the price of calculators dropped, multiplication and exponentiation were performed with an isomor... | ## Note 11.7.11\n\n(b) Any application of this definition requires a procedure outlined in Figure 11.7.10. The first condition, that an isomorphism be a bijection, reflects the fact that every true statement in the first group should have exactly one corresponding true statement in the second group. This is exactly why... | Yes |
Consider \( G = \left\{ {\left. \left( \begin{array}{ll} 1 & a \\ 0 & 1 \end{array}\right) \right| \;a \in \mathbb{R}}\right\} \) with matrix multiplication. The group \( \left\lbrack {\mathbb{R}; + }\right\rbrack \) is isomorphic to \( G \) . | Our translation rule is the function \( f : \mathbb{R} \rightarrow G \) defined by \( f\left( a\right) = \left( \begin{array}{ll} 1 & a \\ 0 & 1 \end{array}\right) \) . Since groups have only one operation, there is no need to state explicitly that addition is translated to matrix multiplication. That \( f \) is a bije... | Yes |
Theorem 12.1.4 Elementary Operations on Equations. If any sequence of the following operations is performed on a system of equations, the resulting system is equivalent to the original system:\n\n(a) Interchange any two equations in the system.\n\n(b) Multiply both sides of any equation by a nonzero constant.\n\n(c) Mu... | Let us now use the above theorem to work out the details of Example 12.1.3 and see how we can arrive at the simpler system.\n\nThe original system:\n\n\[ \n4{x}_{1} + 2{x}_{2} + {x}_{3} = 1 \n\]\n\n\[ \n2{x}_{1} + {x}_{2} + {x}_{3} = 4 \n\]\n\n(12.1.1)\n\n\[ \n2{x}_{1} + 2{x}_{2} + {x}_{3} = 3 \n\]\n\nStep 1. We will f... | No |
Theorem 12.1.5 Elementary Row Operations. If any sequence of the following operations is performed on the augmented matrix of a system of equations, the resulting matrix is a system that is equivalent to the original system. The following operations on a matrix are called elementary row operations:\n\n(1) Exchange any ... | If we use the notation \( {R}_{i} \) to stand for Row \( i \) of a matrix and \( \rightarrow \) to stand for row equivalence, then\n\n\[ A\overset{c{R}_{i} + {R}_{j}}{ \rightarrow }B \]\n\nmeans that the matrix \( B \) is obtained from the matrix \( A \) by multiplying the Row \( i \) of \( A \) by \( c \) and adding t... | Yes |
Find all solutions to the system\n\n\[ \n{x}_{1} + 3{x}_{2} + {x}_{3} = 2 \n\]\n\n\[ \n{x}_{1} + {x}_{2} + 5{x}_{3} = 4 \n\]\n\n\[ \n2{x}_{1} + 2{x}_{2} + {10}{x}_{3} = 6 \n\] | The reader can verify that the augmented matrix of this system, \( \left( \begin{array}{llll} 1 & 3 & 1 & 2 \\ 1 & 1 & 5 & 4 \\ 2 & 2 & {10} & 6 \end{array}\right) \) ,\n\nreduces to \( \left( \begin{matrix} 1 & 3 & 1 & 2 \\ 1 & 1 & 5 & 4 \\ 0 & 0 & 0 & - 2 \end{matrix}\right) \) .\n\nWe can attempt to row-reduce this ... | Yes |
Example 12.1.8 A system with an infinite number of solutions. Next, let's attempt to find all of the solutions to:\n\n\[ \n{x}_{1} + 6{x}_{2} + 2{x}_{3} = 1 \]\n\n\[ \n2{x}_{1} + {x}_{2} + 3{x}_{3} = 2 \]\n\n\[ \n4{x}_{1} + 2{x}_{2} + 6{x}_{3} = 4 \]\n\nThe augmented matrix for the system is\n\n\[ \n\left( \begin{array... | The augmented matrix for the system is\n\n\[ \n\left( \begin{array}{llll} 1 & 6 & 2 & 1 \\ 2 & 1 & 3 & 2 \\ 4 & 2 & 6 & 4 \end{array}\right) \]\n\n(12.1.8)\n\nwhich reduces to\n\n\[ \n\left( \begin{matrix} 1 & 0 & \frac{16}{11} & 1 \\ 0 & 1 & \frac{1}{11} & 0 \\ 0 & 0 & 0 & 0 \end{matrix}\right) \]\n\n(12.1.9)\n\nIf we... | Yes |
If we apply The Gauss-Jordan Algorithm to the system\n\n\\[ \n5{x}_{1} + {x}_{2} + 2{x}_{3} + {x}_{4} = 2 \n\\]\n\n\\[ \n3{x}_{1} + {x}_{2} - 2{x}_{3} = 5 \n\\]\n\n\\[ \n{x}_{1} + {x}_{2} + 3{x}_{3} - {x}_{4} = - 1 \n\\]\n\nthe augmented matrix is\n\n\\[ \n\\left( \\begin{matrix} 5 & 1 & 2 & 1 & 2 \\ 3 & 1 & - 2 & 0 & ... | is reduced to\n\n\\[ \n\\left( \\begin{matrix} 1 & 0 & 0 & \\frac{1}{2} & \\frac{1}{2} \\ 0 & 1 & 0 & - \\frac{3}{2} & \\frac{3}{2} \\ 0 & 0 & 1 & 0 & - 1 \\end{matrix}\\right) \n\\]\n\nTherefore, \\( {x}_{4} \\) is a free variable in the solution and general solution of the system is\n\n\\[ \nx = \\left( \\begin{array... | Yes |
Example 12.2.2 Recognition of a non-invertible matrix. The reader can verify that if \( A = \left( \begin{matrix} 1 & 2 & 1 \\ - 1 & - 2 & - 1 \\ 0 & 5 & 8 \end{matrix}\right) \) then the augmented matrix \( \left( \begin{matrix} 1 & 2 & 1 & 1 & 0 & 0 \\ - 1 & - 2 & - 2 & 0 & 1 & 0 \\ 0 & 5 & 8 & 0 & 0 & 1 \end{matrix}... | \[ \left( \begin{array}{llllll} 1 & 2 & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 & 0 \\ 0 & 5 & 8 & 0 & 0 & 1 \end{array}\right) \] (12.2.4) Although this matrix can be row-reduced further, it is not necessary to do so since, in equation form, we have: Table 12.2.3 \[ {x}_{11} + 2{x}_{21} + {x}_{31} = 1\;{x}_{12} + 2{x}_{22}... | Yes |
A Vector Space of Matrices. Let \( V = {M}_{2 \times 3}\left( \mathbb{R}\right) \) and let the operations of addition and scalar multiplication be the usual operations of addition and scalar multiplication on matrices. Then \( V \) together with these operations is a real vector space. | The reader is strongly encouraged to verify the definition for this example before proceeding further (see Exercise 3 of this section). | No |
The Vector Space \( {\mathbb{R}}^{2} \) . Let \( {\mathbb{R}}^{2} = \left\{ {\left( {{a}_{1},{a}_{2}}\right) \mid {a}_{1},{a}_{2} \in \mathbb{R}}\right\} \) . If we define addition and scalar multiplication the natural way, that is, as we would on \( 1 \times 2 \) matrices, then \( {\mathbb{R}}^{2} \) is a vector space... | See Exercise 12.3.3.4 of this section. | No |
The vector \( \left( {2,3}\right) \) in \( {\mathbb{R}}^{2} \) is a linear combination of the vectors \( \left( {1,0}\right) \) and \( \left( {0,1}\right) \) | since \( \left( {2,3}\right) = 2\left( {1,0}\right) + 3\left( {0,1}\right) \) | Yes |
Prove that the vector \( \left( {4,5}\right) \) is a linear combination of the vectors \( \left( {3,1}\right) \) and \( \left( {1,4}\right) \) . | By the definition we must show that there exist scalars \( {a}_{1} \) and \( {a}_{2} \) such that:\n\n\[ \begin{aligned} \left( {4,5}\right) & = {a}_{1}\left( {3,1}\right) + {a}_{2}\left( {1,4}\right) \\ & = \left( {3{a}_{1} + {a}_{2},{a}_{1} + 4{a}_{2}}\right) \end{aligned}\; \Rightarrow \;\begin{array}{l} 3{a}_{1} + ... | Yes |
Theorem 12.3.15 The fundamental property of a basis. If \( \left\{ {{\mathbf{x}}_{1},{\mathbf{x}}_{2},\ldots ,{\mathbf{x}}_{n}}\right\} \) is a basis for a vector space \( V \) over \( \mathbb{R} \), then any vector \( y \in V \) can be uniquely expressed as a linear combination of the \( {\mathbf{x}}_{i} \)’s. | Proof. Assume that \( \left\{ {{\mathbf{x}}_{1},{\mathbf{x}}_{2},\ldots ,{\mathbf{x}}_{n}}\right\} \) is a basis for \( V \) over \( \mathbb{R} \). We must prove two facts:\n\n(1) each vector \( y \in V \) can be expressed as a linear combination of the \( {\mathbf{x}}_{i} \)’s, and\n\n(2) each such expression is uniqu... | Yes |
Prove that \( \{ \left( {1,1}\right) ,\left( {-1,1}\right) \} \) is a basis for \( {\mathbb{R}}^{2} \) over \( \mathbb{R} \) and explain what this means geometrically. | First we show that the vectors \( \left( {1,1}\right) \) and \( \left( {-1,1}\right) \) generate all of \( {\mathbb{R}}^{2} \) . We can do this by imitating Example 12.3.8 and leave it to the reader (see Exercise 12.3.3.10 of this section). Secondly, we must prove that the set is linearly independent.\n\nLet \( {a}_{1}... | No |
We will now diagonalize the matrix \( A \) of Example 12.4.2. We form the matrix \( P \) as follows: Let \( {P}^{\left( 1\right) } \) be the first column of \( P \) . Choose for \( {P}^{\left( 1\right) } \) any eigenvector from \( {E}_{1} \) . We may as well choose a simple vector in \( {E}_{1} \) so \( {P}^{\left( 1\r... | Similarly, let \( {P}^{\left( 2\right) } \) be the second column of \( P \), and choose for \( {P}^{\left( 2\right) } \) any eigenvector from \( {E}_{2} \) . The vector \( {P}^{\left( 2\right) } = \left( \begin{array}{l} 1 \\ 2 \end{array}\right) \) is a reasonable choice, thus\n\n\[ P = \left( \begin{matrix} 1 & 1 \\ ... | Yes |
Theorem 12.4.9 A condition for diagonalizability. Let \( A \) be an \( n \times n \) matrix. Then \( A \) is diagonalizable if and only if \( A \) has \( n \) linearly independent eigenvectors. | Proof. Outline of a proof: \( \left( \Leftarrow \right) \) Assume that \( A \) has linearly independent eigenvectors, \( {P}^{\left( 1\right) },{P}^{\left( 2\right) },\ldots ,{P}^{\left( n\right) } \), with corresponding eigenvalues \( {\lambda }_{1},{\lambda }_{2},\ldots \) , \( {\lambda }_{n} \) . We want to prove th... | No |
Example 12.4.10 A Matrix that is Not Diagonalizable. Let us attempt to diagonalize the matrix \( A = \left( \begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 1 \\ 1 & - 1 & 4 \end{matrix}\right) \) | First, we determine the eigenvalues.\n\n\[ \det \left( {A - {\lambda I}}\right) = \det \left( \begin{matrix} 1 - \lambda & 0 & 0 \\ 0 & 2 - \lambda & 1 \\ 1 & - 1 & 4 - \lambda \end{matrix}\right) \]\n\n\[ = \left( {1 - \lambda }\right) \det \left( \begin{matrix} 2 - \lambda & 1 \\ - 1 & 4 - \lambda \end{matrix}\right)... | Yes |
Consider the computation of terms of the Fibonnaci Sequence. Recall that \( {F}_{0} = 1,{F}_{1} = 1 \) and \( {F}_{k} = {F}_{k - 1} + {F}_{k - 2} \) for \( k \geq 2 \). | In order to formulate the calculation in matrix form, we introduced the \ | No |
How do we compute \( {A}^{k} \) for possibly large values of \( k \) ? | From the discussion at the beginning of this section, we know that \( {A}^{k} = P{D}^{k}{P}^{-1} \) if \( A \) is diagonalizable. We leave to the reader to show that \( \lambda = 1,2 \), and -1 are eigenvalues of \( A \) with eigenvectors\n\n\[ \left( \begin{matrix} 1 \\ 0 \\ - 1 \end{matrix}\right) ,\left( \begin{arra... | No |
Given a polynomial \( f\left( x\right) \), we defined the matrix-polynomial \( f\left( A\right) \) for square matrices in Chapter 5. Hence, we are in a position to describe \( {e}^{A} \) for an \( n \times n \) matrix \( A \) as a limit of polynomials, the partial sums of the series. Formally, we write\n\n\[ \n{e}^{A} ... | Again we encounter the need to compute high powers of a matrix. Let \( A \) be an \( n \times n \) diagonalizable matrix. Then there exists an invertible \( n \times n \) matrix \( P \) such that \( {P}^{-1}{AP} = D \), a diagonal matrix, so that\n\n\[ \n{e}^{A} = {e}^{{PD}{P}^{-1}} \n\]\n\n\[ \n= \mathop{\sum }\limits... | Yes |
Theorem 13.1.6 Uniqueness of Least Upper and Greatest Lower Bounds. Let \( \left( {L, \preccurlyeq }\right) \) be a poset, and \( a, b \in L \) . If a greatest lower bound of a and \( b \) exists, then it is unique. The same is true of a least upper bound, if it exists. | Proof. Let \( \ell \) and \( {\ell }^{\prime } \) be greatest lower bounds of \( a \) and \( b \) . We will prove that \( \ell = {\ell }^{\prime } \) .\n\n(1) \( \ell \) a greatest lower bound of \( a \) and \( b \Rightarrow \ell \) is a lower bound of \( a \) and \( b \) .\n\n(2) \( {\ell }^{\prime } \) a greatest low... | Yes |
The power set of a three element set. Consider the poset \( \left( {\mathcal{P}\left( A\right) , \subseteq }\right) \), where \( A = \{ 1,2,3\} \) . The greatest lower bound of \( \{ 1,2\} \) and \( \{ 1,3\} \) is \( \ell = \{ 1\} \) . For any other element \( {\ell }^{\prime } \) which is a subset of \( \{ a, b\} \) a... | The least element of \( \mathcal{P}\left( A\right) \) is \( \varnothing \) and the greatest element is \( A = \{ a, b, c\} \) . The Hasse diagram of this poset is shown in Figure 13.1.11. | No |
The power set of a three element set. Consider the poset \( \left( {\mathcal{P}\left( A\right) , \subseteq }\right) \) we examined in Example 13.1.10. It isn’t too surprising that every pair of sets had a greatest lower bound and least upper bound. Thus, we have a lattice in this case; and \( A \vee B = A \cup B \) and... | The reader is encouraged to write out the operation tables \( \left\lbrack {\mathcal{P}\left( A\right) ;\cup , \cap }\right\rbrack \) . | No |
Example 13.2.5 A Nondistributive Lattice. We now give an example of a lattice where the distributive laws do not hold. Let \( L = \{ \mathbf{0}, a, b, c,\mathbf{1}\} \) . We define the partial ordering \( \preccurlyeq \) on \( L \) by the set\n\n\[ \n\{ \left( {\mathbf{0},\mathbf{0}}\right) ,\left( {\mathbf{0}, a}\righ... | We note that: \( a \vee \left( {c \land b}\right) = \) \( a \vee \mathbf{0} = a \) and \( \left( {a \vee c}\right) \land \left( {a \vee b}\right) = \mathbf{1} \land \mathbf{1} = \mathbf{1} \) . Therefore, \( a \vee \left( {b \land c}\right) \neq \left( {a \vee b}\right) \land \left( {a \vee c}\right) \) for some values... | Yes |
Set Complement is a Complement. In Chapter 1, we defined the complement of a subset of any universe. This turns out to be a concrete example of the general concept we have just defined, but we will reason through why this is the case here. Let \( L = \mathcal{P}\left( A\right) \), where \( A = \{ a, b, c\} \). Then \( ... | It’s not too difficult to see that \( D = \{ c\} \), since we need to include \( c \) to make the first condition true and can’t include \( a \) or \( b \) if the second condition is to be true. Of course this is precisely how we defined \( {A}^{c} \) in Chapter 1. Since it can be shown that each element of \( L \) has... | No |
Theorem 13.3.7 One condition for unique complements. If \( \left\lbrack {L;\vee , \land }\right\rbrack \) is a complemented, distributive lattice, then the complement of each element \( a \in L \) is unique. | Proof. Let \( a \in L \) and assume to the contrary that \( a \) has two complements, namely \( {a}_{1} \) and \( {a}_{2} \) . Then by the definition of complement,\n\n\[ \begin{matrix} a \land {a}_{1} = 0\text{ and }a \vee {a}_{1} = 1, \\ \text{ and } \end{matrix} \]\n\n\[ a \land {a}_{2} = 0\text{and}a \vee {a}_{2} =... | Yes |
Theorem 13.4.6 Let \( \mathcal{B} = \left\lbrack {B;\vee ,\land , - }\right\rbrack \) be any finite Boolean algebra, and let \( A \) be the set of all atoms of \( \mathcal{B} \) . Then \( \left\lbrack {\mathcal{P}\left( A\right) ;\cup ,\cap ,{}^{c}}\right\rbrack \) is isomorphic to \( \left\lbrack {B;\vee ,\land , - }\... | Proof. An isomorphism that serves to prove this theorem is \( T : \mathcal{P}\left( A\right) \rightarrow B \) defined by \( T\left( S\right) = \mathop{\bigvee }\limits_{{a \in S}}a \), where \( T\left( \varnothing \right) \) is interpreted as the zero of \( \mathcal{B} \) . We leave it to the reader to prove that this ... | No |
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