A First Course in Linear Algebra, 2017a

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502 Vector Spaces Example 9.68: Onto and Spanning Let V and W be vector spaces and T : V → W a linear transformation. Prove that if T is onto and V = span{⃗v 1 ,⃗v 2 ,...,⃗v k },then W = span{T (⃗v 1 ),T (⃗v 2 ),...,T (⃗v k )}. Solution. Suppose that T is onto and let ⃗w ∈ W. Then there exists ⃗v ∈ V such that T (⃗v) =⃗w. Since V = span{⃗v 1 ,⃗v 2 ,...,⃗v k },thereexista 1 ,a 2 ,...a k ∈ R such that⃗v = a 1 ⃗v 1 + a 2 ⃗v 2 + ···+ a k ⃗v k . Using the fact that T is a linear transformation, ⃗w = T (⃗v) = T (a 1 ⃗v 1 + a 2 ⃗v 2 + ···+ a k ⃗v k ) = a 1 T (⃗v 1 )+a 2 T (⃗v 2 )+···+ a k T (⃗v k ), i.e., ⃗w ∈ span{T(⃗v 1 ),T(⃗v 2 ),...,T (⃗v k )}, and thus W ⊆ span{T (⃗v 1 ),T (⃗v 2 ),...,T (⃗v k )}. Since T (⃗v 1 ),T (⃗v 2 ),...,T (⃗v k ) ∈ W, it follows from that span{T (⃗v 1 ),T (⃗v 2 ),...,T (⃗v k )}⊆W, and therefore W = span{T (⃗v 1 ),T (⃗v 2 ),...,T (⃗v k )}. ♠ 9.7.2 Isomorphisms The focus of this section is on linear transformations which are both one to one and onto. When this is the case, we call the transformation an isomorphism. Definition 9.69: Isomorphism Let V and W be two vector spaces and let T : V ↦→ W be a linear transformation. Then T is called an isomorphism if the following two conditions are satisfied. • T is one to one. • T is onto. Definition 9.70: Isomorphic Let V and W be two vector spaces and let T : V ↦→ W be a linear transformation. Then if T is an isomorphism, we say that V and W are isomorphic. Consider the following example of an isomorphism.
9.7. Isomorphisms 503 Example 9.71: Isomorphism Let T : M 22 → R 4 be defined by [ a b T c d ⎡ ] = ⎢ ⎣ a b c d ⎤ [ ⎥ a ⎦ for all c b d ] ∈ M 22 . Show that T is an isomorphism. Solution. Notice that if we can prove T is an isomorphism, it will mean that M 22 and R 4 are isomorphic. It remains to prove that 1. T is a linear transformation; 2. T is one-to-one; 3. T is onto. T is linear: Let k, p be scalars. ( [ ] [ ]) a1 b T k 1 a2 b + p 2 c 1 d 1 c 2 d 2 ([ ] ka1 kb = T 1 kc 1 kd 1 = T ⎡ = = ⎢ ⎣ ⎡ ⎢ ⎣ = k ⎢ ⎣ [ ]) pa2 pb + 2 pc 2 pd 2 ([ ]) ka1 + pa 2 kb 1 + pb 2 kc 1 + pc 2 kd 1 + pd 2 ⎤ ka 1 + pa 2 kb 1 + pb 2 ⎥ kc 1 + pc 2 ⎦ kd 1 + pd 2 ⎤ ⎡ ka 1 kb 1 ⎥ kc 1 ⎦ + ⎢ ⎣ kd 1 ⎡ = kT ⎤ a 1 b 1 ⎥ c 1 d 1 ⎡ ⎦ + p ⎢ ⎣ ⎤ pa 2 pb 2 ⎥ pc 2 ⎦ pd 2 ⎤ a 2 b 2 ⎥ c 2 ⎦ d 2 ([ ]) a1 b 1 + pT c 1 d 1 ([ ]) a2 b 2 c 2 d 2 Therefore T is linear. T is one-to-one: By Lemma 9.64 we need to show that if T (A) =0thenA = 0 for some matrix A ∈ M 22 . ⎡ ⎤ ⎡ ⎤ [ ] a 0 a b T = ⎢ b ⎥ c d ⎣ c ⎦ = ⎢ 0 ⎥ ⎣ 0 ⎦ d 0
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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
Page 465 and 466: 9.1. Algebraic Considerations 451 E
Page 467 and 468: 9.1. Algebraic Considerations 453 H
Page 469 and 470: 9.1. Algebraic Considerations 455
Page 471 and 472: 9.1. Algebraic Considerations 457 =
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Page 485 and 486: 9.3. Linear Independence 471 Soluti
Page 487 and 488: 9.3. Linear Independence 473 Exerci
Page 489 and 490: 9.3. Linear Independence 475 Exerci
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Page 493 and 494: 9.4. Subspaces and Basis 479 1. U
Page 495 and 496: 9.4. Subspaces and Basis 481 It fol
Page 497 and 498: 9.4. Subspaces and Basis 483 But th
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Page 501 and 502: 9.4. Subspaces and Basis 487 Then f
Page 503 and 504: 9.4. Subspaces and Basis 489 Theref
Page 505 and 506: 9.4. Subspaces and Basis 491 The re
Page 507 and 508: 9.6. Linear Transformations 493 2.
Page 509 and 510: 9.6. Linear Transformations 495 2.
Page 511 and 512: 9.6. Linear Transformations 497 The
Page 513 and 514: 9.7. Isomorphisms 499 9.7 Isomorphi
Page 515: 9.7. Isomorphisms 501 Example 9.66:
Page 519 and 520: 9.7. Isomorphisms 505 Since T is on
Page 521 and 522: 9.7. Isomorphisms 507 for ∑ n i=1
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 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
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
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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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A First Course in Linear Algebra, 2017a

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