A First Course in Linear Algebra, 2017a

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534 Some Prerequisite Topics Suppose then that the inequality holds for n.Inotherwords, 1 2 · 3 ···2n − 1 1 < √ 4 2n 2n + 1 is true. Now multiply both sides of this inequality by 2n+1 2n+2 . This yields 1 2 · 3 ···2n − 1 · 2n + 1 √ 4 2n 2n + 2 < 1 2n + 1 2n + 1 √ 2n + 1 2n + 2 = 2n + 2 The theorem will be proved if this last expression is less than ( ) 1 2 √ = 1 2n + 3 2n + 3 > 2n + 1 (2n + 2) 2 1 √ 2n + 3 . This happens if and only if which occurs if and only if (2n + 2) 2 > (2n + 3)(2n + 1) and this is clearly true which may be seen from expanding both sides. This proves the inequality. ♠ Let’s review the process just used. If S is the set of integers at least as large as 1 for which the formula holds, the first step was to show 1 ∈ S and then that whenever n ∈ S, itfollowsn + 1 ∈ S. Therefore, by the principle of mathematical induction, S contains [1,∞) ∩ Z, all positive integers. In doing an inductive proof of this sort, the set S is normally not mentioned. One just verifies the steps above.
Appendix B Selected Exercise Answers 1.1.1 x + 3y = 1 4x − y = 3 , Solution is: [ x = 10 13 ,y = 1 13] . 1.1.2 3x + y = 3 x + 2y = 1 , Solution is: [x = 1,y = 0] 1.2.1 x + 3y = 1 4x − y = 3 , Solution is: [ x = 10 13 ,y = 13 1 ] 1.2.2 x + 3y = 1 4x − y = 3 , Solution is: [ x = 10 13 ,y = 13 1 ] 1.2.3 x + 2y = 1 2x − y = 1 4x + 3y = 3 , Solution is: [ x = 3 5 ,y = 1 ] 5 1.2.4 ⎡ No solution⎤ exists. You can see this ⎡ by writing the ⎤ augmented matrix and doing row operations. 1 1 −3 2 1 0 4 0 ⎣ 2 1 1 1 ⎦, row echelon form: ⎣ 0 1 −7 0 ⎦. Thus one of the equations says 0 = 1inan 3 2 −2 0 0 0 0 1 equivalent system of equations. 1.2.5 4g − I = 150 4I − 17g = −660 4g + s = 290 g + I + s − b = 0 , Solution is : {g = 60,I = 90,b = 200,s = 50} 1.2.6 The solution exists but is not unique. 1.2.7 A solution exists and is unique. 1.2.9 There might be a solution. If so, there are infinitely many. 1.2.10 No. Consider x + y + z = 2andx + y + z = 1. 1.2.11 These can have a solution. For example, x + y = 1,2x + 2y = 2,3x + 3y = 3evenhasaninfiniteset of solutions. 1.2.12 h = 4 1.2.13 Any h will work. 535
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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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324 Complex Numbers Theorem 6.1: Pr
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326 Complex Numbers Example 6.5: Co
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328 Complex Numbers Definition 6.10
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330 Complex Numbers Exercise 6.1.5
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332 Complex Numbers Example 6.14: P
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334 Complex Numbers This requires r
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336 Complex Numbers Therefore, x 3
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338 Complex Numbers Consider the fo
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Chapter 7 Spectral Theory 7.1 Eigen
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7.1. Eigenvalues and Eigenvectors o
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7.2. Diagonalization 355 One eigenv
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7.2. Diagonalization 357 In words,
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7.2. Diagonalization 359 Conversely
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7.2. Diagonalization 361 Theorem 7.
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7.2. Diagonalization 363 7.2.3 Comp
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7.2. Diagonalization 365 Exercise 7
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7.3. Applications of Spectral Theor
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7.4. Orthogonality 395 [ x(0) y(0)
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7.4. Orthogonality 397 Solution. Fi
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7.4. Orthogonality 399 Therefore an
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7.4. Orthogonality 401 In the above
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7.4. Orthogonality 403 Theorem 7.62
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7.4. Orthogonality 405 Theorem 7.66
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7.4. Orthogonality 407 λ 1 = 16: s
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7.4. Orthogonality 409 [ ] [ ] AV1
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7.4. Orthogonality 411 = [ 4 0 0 0
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7.4. Orthogonality 413 by the assum
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7.4. Orthogonality 415 operations)
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7.4. Orthogonality 417 Notice that
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7.4. Orthogonality 419 insignifican
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7.4. Orthogonality 421 Usually it w
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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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Page 531 and 532: 9.9. The Matrix of a Linear Transfo
Page 543 and 544: Appendix A Some Prerequisite Topics
Page 545 and 546: A.2. Well Ordering and Induction 53
Page 547: A.2. Well Ordering and Induction 53
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Page 553 and 554: 539 2.1.7 0A =(0 + 0)A = 0A + 0A.No
Page 555 and 556: 541 [ 2.1.18 Let A = 2.1.20 Let A =
Page 557 and 558: 543 2.1.40 2.1.41 ⎡ ⎣ ⎡ ⎣ 1
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Page 561 and 562: 547 3.1.14 If the determinant is no
Page 563 and 564: 549 ⎡ 3.2.2 det⎣ 1 2 0 0 2 1 3
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Page 569 and 570: 555 It follows that the expression
Page 571 and 572: 557 4.11.6 (a) Orthogonal. (b) Symm
Page 573 and 574: 559 4.11.14 Then a solution is ⎡
Page 575 and 576: 561 4.12.23 ⎡ ⎢ ⃗F 1 • ⎣
Page 577 and 578: 563 5.2.13 ⎡ 1 ⎣ 10 1 0 3 0 0 0
Page 579 and 580: 565 Now to write in terms of (a,b),
Page 581 and 582: 567 ⎡ 5.9.7 Solution is: ⎣ −
Page 583 and 584: 569 6.1.6 If p(z)=0, then you have
Page 585 and 586: 571 7.1.2 Say AX = λX.ThencAX = c
Page 587 and 588: 573 7.3.1 First we write A = PDP
Page 589 and 590: 575 7.3.19 The solution is [ e At 8
Page 591 and 592: 577 7.4.12 The eigenvectors and eig
Page 593 and 594: 579 9.3.30 No. They can’t be. 9.3
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Page 597 and 598: Index ∩, 529 ∪, 529 \, 529 row-
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INDEX 585 null space, 211 orthogona
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A First Course in Linear Algebra, 2017a

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