A First Course in Linear Algebra, 2017a

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374 Spectral Theory = ⎡ ⎣ 102 164 434 ⎤ ⎦ We could progress in this manner to find the populations after 10 time periods. However from our above discussion, we can simply calculate (A n X 0 ) i ,wheren denotes the number of time periods which have passed. Therefore, we compute the populations in each location after 10 units of time as follows. ⎡ ⎣ X 10 = A 10 X 0 ⎡ ⎤ x 110 x 210 ⎦ = x 310 = ⎣ ⎡ ⎣ .6 0 .1 .2 .8 0 .2 .2 .9 ⎤ ⎦ 10 ⎡ ⎣ 115.08582922 120.13067244 464.78349834 100 200 400 ⎤ Since we are speaking about populations, we would need to round these numbers to provide a logical answer. Therefore, we can say that after 10 units of time, there will be 115 residents in location one, 120 in location two, and 465 in location three. ♠ A second important application of Markov matrices is the concept of random walks. Suppose a walker has m locations to choose from, denoted 1,2,···,m. Let a ij refer to the probability that the person will travel to location i from location j. Again, this requires that k ∑ i=1 a ij = 1 In this context, the vector X n =[x 1n ···x mn ] T contains the probabilities x in the walker ends up in location i,1≤ i ≤ m at time n. Example 7.35: Random Walks Suppose three locations exist, referred to as locations 1,2 and 3. The Markov matrix of probabilities A =[a ij ] is given by ⎡ ⎤ 0.4 0.1 0.5 ⎣ 0.4 0.6 0.1 ⎦ 0.2 0.3 0.4 If the walker starts in location 1, calculate the probability that he ends up in location 3 at time n = 2. ⎦ ⎤ ⎦ Solution. Since the walker begins in location 1, we have ⎡ ⎤ 1 X 0 = ⎣ 0 ⎦ 0 The goal is to calculate x 32 . To do this we calculate X 2 ,usingX n+1 = AX n . X 1 = AX 0
7.3. Applications of Spectral Theory 375 = = ⎡ ⎣ ⎡ ⎣ 0.4 0.1 0.5 0.4 0.6 0.1 0.2 0.3 0.4 0.4 0.4 0.2 ⎤ ⎦ ⎤⎡ ⎦⎣ 1 0 0 ⎤ ⎦ X 2 = AX ⎡ 1 ⎤⎡ 0.4 0.1 0.5 = ⎣ 0.4 0.6 0.1 ⎦⎣ 0.2 0.3 0.4 ⎡ ⎤ 0.3 = ⎣ 0.42 ⎦ 0.28 0.4 0.4 0.2 ⎤ ⎦ This gives the probabilities that our walker ends up in locations 1, 2, and 3. For this example we are interested in location 3, with a probability on 0.28. ♠ Returning to the context of migration, suppose we wish to know how many residents will be in a certain location after a very long time. It turns out that if some power of the migration matrix has all positive entries, then there is a vector X s such that A n X 0 approaches X s as n becomes very large. Hence as more time passes and n increases, A n X 0 will become closer to the vector X s . Consider Theorem 7.33. Letn increase so that X n approaches X s .AsX n becomes closer to X s ,sotoo does X n+1 . For sufficiently large n, the statement X n+1 = AX n can be written as X s = AX s . This discussion motivates the following theorem. Theorem 7.36: Steady State Vector Let A be a migration matrix. Then there exists a steady state vector written X s such that X s = AX s where X s has positive entries which have the same sum as the entries of X 0 . As n increases, the state vectors X n will approach X s . Note that the condition in Theorem 7.36 can be written as (I − A)X s = 0, representing a homogeneous system of equations. Consider the following example. Notice that it is the same example as the Example 7.34 but here it will involve a longer time frame.
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with Open Texts Linear A Algebra wi
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A First Course in Linear Algebra an
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A First Course in Linear Algebra an
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iv Table of Contents 3 Determinants
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vi Table of Contents 7.3.5 The Matr
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Chapter 1 Systems of Equations 1.1
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1.1. Systems of Equations, Geometry
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1.2. Systems of Equations, Algebrai
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54 Matrices Consider the following
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56 Matrices Solution. Notice that b
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58 Matrices Proposition 2.10: Prope
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60 Matrices on the left by a vector
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62 Matrices will satisfy the equati
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64 Matrices Solution. First check i
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66 Matrices Solution. First check i
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68 Matrices Proposition 2.25: Prope
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70 Matrices Definition 2.29: Symmet
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72 Matrices Theorem 2.34: Uniquenes
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74 Matrices Let’s solve the first
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76 Matrices Notice that the left ha
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78 Matrices What if the right side,
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80 Matrices First, consider the fol
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82 Matrices You can see that B is t
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84 Matrices = = ⎡ ⎣ ⎡ ⎣ 1 0
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86 Matrices First: ⎡ ⎣ 0 1 0 1
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88 Matrices Proof. Suppose that A a
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90 Matrices It follows that the red
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92 Matrices (h) DE ⎡ Exercise 2.1
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94 Matrices (a) (A − B) 2 = A 2
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96 Matrices Exercise 2.1.42 Let ⎡
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98 Matrices (a) Find the elementary
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100 Matrices One way to find the LU
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102 Matrices which yields very quic
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104 Matrices By Lemma 2.70, this im
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106 Matrices Exercise 2.2.12 Find t
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108 Determinants The 2×2 determina
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110 Determinants Definition 3.7: Th
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112 Determinants The following prov
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114 Determinants 3.1.3 Properties o
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116 Determinants Example 3.23: Mult
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118 Determinants Solution. Consider
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120 Determinants = ∑a 1,l (cof(B)
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122 Determinants Theorem 3.35: Let
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124 Determinants Here, det(C)=−3d
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126 Determinants (a) minor(A) 11 (b
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128 Determinants Exercise 3.1.13 An
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130 Determinants 3.2.1 A Formula fo
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132 Determinants You can verify tha
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134 Determinants The cofactor matri
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136 Determinants Therefore, ∣ ∣
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138 Determinants After row operatio
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140 Determinants Using the given po
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142 Determinants Does there exist a
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144 R n Hence, every element in R 2
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146 R n components are equal. Preci
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148 R n Theorem 4.6: Properties of
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150 R n 4.3 Geometric Meaning of Ve
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152 R n 4.4 Length of a Vector Outc
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154 R n Solution. We will use the f
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156 R n = √ 1 + 9 + 16 = √ 26 I
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158 R n ⃗u 2⃗v ⃗u + 2⃗v ⃗
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160 R n ⎡ Let ⃗q = ⎣ Then, x
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162 R n Let t = x−2 y−1 3 ,t =
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164 R n Example 4.26: Compute a Dot
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166 R n Notice that this proof was
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168 R n Therefore, the cosine of th
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170 R n 4.7.3 Projections In some a
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172 R n We will conclude this secti
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174 R n Exercise 4.7.11 Find proj
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176 R n Solution. By Definition 4.4
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178 R n From the above scalar equat
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180 R n Definition 4.48: Geometric
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182 R n We will now look at an exam
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184 R n 4.9.1 The Box Product Recal
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186 R n = a ∣ e h ⎡ = det⎣ f
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188 R n 4.10 Spanning, Linear Indep
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190 R n 4.10.2 Linearly Independent
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192 R n Theorem 4.66: Linear Indepe
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194 R n Therefore we can write ⎡
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196 R n Now suppose that⃗u ∉ sp
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198 R n A subspace is simply a set
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200 R n Definition 4.80: Basis of a
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202 R n ⃗u 1 = Furthermore, ⎡
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204 R n ⎡ ⎢ ⎣ 1 2 1 1 3 0 1 3
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206 R n Solution. An easy way to do
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208 R n Since {⃗r 1 ,..., p⃗r j
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210 R n Example 4.101: Rank, Column
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212 R n Definition 4.106: Image of
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214 R n Example 4.110: Null Space o
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216 R n Theorem 4.114: Let A be an
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218 R n Exercise 4.10.7 Here are so
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220 R n Exercise 4.10.17 Are the fo
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222 R n Thse vectors can’t possib
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224 R n ⎧⎡ ⎪⎨ Exercise 4.10
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226 R n Exercise 4.10.52 Consider t
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228 R n Exercise 4.10.68 Find the r
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230 R n Solution. We already verifi
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232 R n Thus to find a correspondin
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234 R n That is: We readily compute
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236 R n We will say that the column
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238 R n To show that {⃗v 1 ,··
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240 R n is an orthogonal basis of U
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242 R n = = = ⎡ ⎣ ⎡ ⎣ ⎡
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244 R n In order to find W ⊥ , we
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246 R n Now, we need to write ⃗y
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248 R n Theorem 4.150: Existence of
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250 R n Solution. First, consider w
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252 R n −1 1 2 3 4 5 6 y 4 2 x Re
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254 R n Exercise 4.11.7 For U an or
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256 R n and here is a point: (1,1,2
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258 R n What does addition of vecto
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260 R n 4 3 First we want to know t
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262 R n = −58 Newton meters Note
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264 R n Exercise 4.12.19 A girl dra
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266 Linear Transformations numerica
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268 Linear Transformations Exercise
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270 Linear Transformations In this
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272 Linear Transformations ⎡ ⎣
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274 Linear Transformations ⎡ T
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276 Linear Transformations ⎡ (b)
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278 Linear Transformations The nece
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280 Linear Transformations Then, co
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282 Linear Transformations More gen
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284 Linear Transformations Solution
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288 Linear Transformations Definiti
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290 Linear Transformations This tel
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292 Linear Transformations Example
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294 Linear Transformations 2. T is
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296 Linear Transformations Proposit
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298 Linear Transformations Proof. S
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300 Linear Transformations You can
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304 Linear Transformations for exam
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306 Linear Transformations Example
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308 Linear Transformations Since {T
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310 Linear Transformations Find a b
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312 Linear Transformations Theorem
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314 Linear Transformations and one
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316 Linear Transformations Hence
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318 Linear Transformations system.
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320 Linear Transformations Solution
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Page 338 and 339: 324 Complex Numbers Theorem 6.1: Pr
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Page 371 and 372: 7.2. Diagonalization 357 In words,
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Page 375 and 376: 7.2. Diagonalization 361 Theorem 7.
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Page 409 and 410: 7.4. Orthogonality 395 [ x(0) y(0)
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Page 437 and 438: 7.4. Orthogonality 423 We now turn
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7.4. Orthogonality 425 Solution. No
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7.4. Orthogonality 427 = 2y 2 1 + 8
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7.4. Orthogonality 429 ⎡ A = ⎢
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7.4. Orthogonality 431 Exercise 7.4
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Chapter 8 Some Curvilinear Coordina
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8.1. Polar Coordinates and Polar Gr
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8.2. Spherical and Cylindrical Coor
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Chapter 9 Vector Spaces 9.1 Algebra
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9.1. Algebraic Considerations 451 E
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9.1. Algebraic Considerations 453 H
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9.1. Algebraic Considerations 455
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9.1. Algebraic Considerations 457 =
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9.2. Spanning Sets 465 Show that, w
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9.2. Spanning Sets 467 Clearlynoval
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9.3. Linear Independence 469 9.3 Li
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9.3. Linear Independence 471 Soluti
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9.3. Linear Independence 473 Exerci
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9.3. Linear Independence 475 Exerci
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9.4. Subspaces and Basis 477 (b) {
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9.4. Subspaces and Basis 479 1. U
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9.4. Subspaces and Basis 481 It fol
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9.4. Subspaces and Basis 483 But th
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9.4. Subspaces and Basis 485 Then {
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9.4. Subspaces and Basis 487 Then f
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9.4. Subspaces and Basis 489 Theref
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9.4. Subspaces and Basis 491 The re
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9.6. Linear Transformations 493 2.
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9.6. Linear Transformations 495 2.
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9.6. Linear Transformations 497 The
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9.7. Isomorphisms 499 9.7 Isomorphi
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9.7. Isomorphisms 501 Example 9.66:
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9.7. Isomorphisms 503 Example 9.71:
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9.7. Isomorphisms 505 Since T is on
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9.7. Isomorphisms 507 for ∑ n i=1
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9.7. Isomorphisms 509 Exercises Exe
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9.7. Isomorphisms 511 for example.
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9.8. The Kernel And Image Of A Line
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9.8. The Kernel And Image Of A Line
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9.9. The Matrix of a Linear Transfo
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Appendix A Some Prerequisite Topics
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A.2. Well Ordering and Induction 53
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A.2. Well Ordering and Induction 53
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Appendix B Selected Exercise Answer
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537 1.2.33 Solution is: [x = 2 −
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539 2.1.7 0A =(0 + 0)A = 0A + 0A.No
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541 [ 2.1.18 Let A = 2.1.20 Let A =
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543 2.1.40 2.1.41 ⎡ ⎣ ⎡ ⎣ 1
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545 2.2.10 An LU factorization of t
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547 3.1.14 If the determinant is no
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549 ⎡ 3.2.2 det⎣ 1 2 0 0 2 1 3
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551 3.2.16 This follows because det
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553 4.7.16 ( ) ( ) ⃗u •⃗v ⃗
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555 It follows that the expression
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557 4.11.6 (a) Orthogonal. (b) Symm
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559 4.11.14 Then a solution is ⎡
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561 4.12.23 ⎡ ⎢ ⃗F 1 • ⎣
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563 5.2.13 ⎡ 1 ⎣ 10 1 0 3 0 0 0
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565 Now to write in terms of (a,b),
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567 ⎡ 5.9.7 Solution is: ⎣ −
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569 6.1.6 If p(z)=0, then you have
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571 7.1.2 Say AX = λX.ThencAX = c
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573 7.3.1 First we write A = PDP
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575 7.3.19 The solution is [ e At 8
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577 7.4.12 The eigenvectors and eig
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579 9.3.30 No. They can’t be. 9.3
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581 9.8.5 We can easily see that di
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Index ∩, 529 ∪, 529 \, 529 row-
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INDEX 585 null space, 211 orthogona
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