A First Course in Linear Algebra, 2017a

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542 Selected Exercise Answers (c) [ −7 1 5 ] [ ] 1 3 (d) 3 9 ⎡ 13 −16 ⎤ 1 (e) ⎣ −16 29 −8 ⎦ 1 −8 5 (f) [ 5 7 ] −1 5 15 5 (g) Not possible. 2.1.28 Show that 1 ( 2 A T + A ) is symmetric and then consider using this as one of the matrices. A = A+A T 2 + A−AT 2 . 2.1.29 If A is symmetric then A = −A T .Itfollowsthata ii = −a ii and so each a ii = 0. 2.1.31 (I m A) ij ≡ ∑ j δ ik A kj = A ij 2.1.32 Yes B = C. Multiply AB = AC on the left by A −1 . ⎡ 2.1.34 A = ⎣ 1 0 0 0 −1 0 0 0 1 ⎤ ⎦ 2.1.35 [ 2 1 −1 3 ⎡ ] −1 = ⎣ 3 7 − 1 ⎤ 7 ⎦ 1 2 7 7 2.1.36 2.1.37 [ 0 1 5 3 [ 2 1 3 0 ] −1 = ] −1 = [ − 3 5 1 5 1 0 [ 0 1 3 1 − 2 3 ] ] 2.1.38 [ 2 1 4 2 ] −1 does not exist. The reduced row-echelon form of this matrix is [ 1 1 2 0 0 ] 2.1.39 [ a b c d ] −1 = [ d ad−bc − ad−bc b ad−bc c a ad−bc − ]
543 2.1.40 2.1.41 ⎡ ⎣ ⎡ ⎣ 1 2 3 2 1 4 1 0 2 1 0 3 2 3 4 1 0 2 ⎤ ⎦ ⎤ ⎦ −1 −1 ⎡ = ⎣ = ⎡ ⎢ ⎣ −2 4 −5 0 1 −2 1 −2 3 −2 0 3 0 1 3 − 2 3 1 0 −1 2.1.42 The reduced row-echelon form is 2.1.43 ⎡ ⎢ ⎣ 1 2 0 2 1 1 2 0 2 1 −3 2 1 2 1 2 ⎤ ⎥ ⎦ −1 ⎡ ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ ⎤ ⎦ 1 0 5 3 0 1 2 3 0 0 0 −1 1 2 1 2 1 2 ⎤ 1 3 2 − 1 2 − 5 2 = ⎢ −1 0 0 1 ⎥ ⎣ −2 − 3 1 9 ⎦ 4 4 4 ⎥ ⎦. There is no inverse. ⎤ 2.1.45 (a) (b) ⎡ ⎣ ⎡ ⎣ x y z x y z ⎤ ⎤ ⎡ ⎣ x y z ⎡ ⎦ = ⎣ ⎡ ⎦ = ⎣ ⎤ ⎡ 1 ⎦ = ⎣ − 2 3 0 ⎤ −12 1 ⎦ 5 3c − 2a 1 3 b − 2 3 c a − c ⎤ ⎦ ⎤ ⎦ 2.1.46 Multiply both sides of AX = B on the left by A −1 . 2.1.47 Multiply on both sides on the left by A −1 . Thus 0 = A −1 0 = A −1 (AX)= ( A −1 A ) X = IX = X 2.1.48 A −1 = A −1 I = A −1 (AB)= ( A −1 A ) B = IB = B. 2.1.49 You need to show that ( A −1) T actsliketheinverseofA T because from uniqueness in the above problem, this will imply it is the inverse. From properties of the transpose, A T ( A −1) T = ( A −1 A ) T = I T = I ( A −1 ) T A T = ( AA −1) T = I T = I Hence ( A −1) T = ( A T ) −1 and this last matrix exists.
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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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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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Page 519 and 520: 9.7. Isomorphisms 505 Since T is on
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Page 523 and 524: 9.7. Isomorphisms 509 Exercises Exe
Page 525 and 526: 9.7. Isomorphisms 511 for example.
Page 527 and 528: 9.8. The Kernel And Image Of A Line
Page 529 and 530: 9.8. The Kernel And Image Of A Line
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 and 548: A.2. Well Ordering and Induction 53
Page 549 and 550: Appendix B Selected Exercise Answer
Page 551 and 552: 537 1.2.33 Solution is: [x = 2 −
Page 553 and 554: 539 2.1.7 0A =(0 + 0)A = 0A + 0A.No
Page 555: 541 [ 2.1.18 Let A = 2.1.20 Let A =
Page 559 and 560: 545 2.2.10 An LU factorization of t
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
Page 565 and 566: 551 3.2.16 This follows because det
Page 567 and 568: 553 4.7.16 ( ) ( ) ⃗u •⃗v ⃗
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
Page 595 and 596: 581 9.8.5 We can easily see that di
Page 597 and 598: Index ∩, 529 ∪, 529 \, 529 row-
Page 599 and 600: INDEX 585 null space, 211 orthogona
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