A First Course in Linear Algebra, 2017a

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458 Vector Spaces • Finally, we show that 1p(x)=p(x). a(bp(x)) = = a ( b ( p 2 x 2 )) + p 1 x + p 0 a ( bp 2 x 2 ) + bp 1 x + bp 0 = = abp 2 x 2 + abp 1 x + abp 0 (ab) ( p 2 x 2 ) + p 1 x + p 0 = (ab)p(x) 1p(x) = 1 ( p 2 x 2 ) + p 1 x + p 0 = 1p 2 x 2 + 1p 1 x + 1p 0 = p 2 x 2 + p 1 x + p 0 = p(x) Since the above axioms hold, we know that P 2 as described above is a vector space. ♠ Another important example of a vector space is the set of all matrices of the same size. Example 9.6: Vector Space of Matrices Let M 2,3 be the set of all 2 × 3 matrices. Using the usual operations of matrix addition and scalar multiplication, show that M 2,3 is a vector space. Solution. Let A,B be 2 × 3 matrices in M 2,3 . We first prove the axioms for addition. • In order to prove that M 2,3 is closed under matrix addition, we show that the sum A + B is in M 2,3 . This means showing that A + B is a 2 × 3matrix. [ ] [ ] a11 a A + B = 12 a 13 b11 b + 12 b 13 a 21 a 22 a 23 b 21 b 22 b 23 [ ] a11 + b = 11 a 12 + b 12 a 13 + b 13 a 21 + b 21 a 22 + b 22 a 23 + b 23 You can see that the sum is a 2×3 matrix, so it is in M 2,3 . It follows that M 2,3 is closed under matrix addition. • The remaining axioms regarding matrix addition follow from properties of matrix addition. Therefore M 2,3 satisfies the axioms of matrix addition. We now turn our attention to the axioms regarding scalar multiplication. Let A,B be matrices in M 2,3 and let c be a real number.
9.1. Algebraic Considerations 459 • We first show that M 2,3 is closed under scalar multiplication. That is, we show that cA a2×3matrix. [ ] a11 a cA = c 12 a 13 a 21 a 22 a 23 [ ] ca11 ca = 12 ca 13 ca 21 ca 22 ca 23 This is a 2 × 3matrixinM 2,3 which proves that the set is closed under scalar multiplication. • The remaining axioms of scalar multiplication follow from properties of scalar multiplication of matrices. Therefore M 2,3 satisfies the axioms of scalar multiplication. In conclusion, M 2,3 satisfies the required axioms and is a vector space. ♠ While here we proved that the set of all 2 × 3 matrices is a vector space, there is nothing special about this choice of matrix size. In fact if we instead consider M m,n ,thesetofallm × n matrices, then M m,n is a vector space under the operations of matrix addition and scalar multiplication. We now examine an example of a set that does not satisfy all of the above axioms, and is therefore not a vector space. Example 9.7: Not a Vector Space Let V denote the set of 2 × 3 matrices. Let addition in V be defined by A + B = A for matrices A,B in V. Let scalar multiplication in V be the usual scalar multiplication of matrices. Show that V is not a vector space. Solution. In order to show that V is not a vector space, it suffices to find only one axiom which is not satisfied. We will begin by examining the axioms for addition until one is found which does not hold. Let A,B be matrices in V . • We first want to check if addition is closed. Consider A + B. By the definition of addition in the example, we have that A + B = A. SinceA is a 2 × 3 matrix, it follows that the sum A + B is in V, and V is closed under addition. • We now wish to check if addition is commutative. That is, we want to check if A + B = B + A for all choices of A and B in V . From the definition of addition, we have that A + B = A and B + A = B. Therefore, we can find A, B in V such that these sums are not equal. One example is A = [ 1 0 0 0 0 0 Using the operation defined by A + B = A, wehave ] ,B = [ 0 0 0 1 0 0 A + B = A [ ] 1 0 0 = 0 0 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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Page 437 and 438: 7.4. Orthogonality 423 We now turn
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Page 445 and 446: 7.4. Orthogonality 431 Exercise 7.4
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Page 463 and 464: 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
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Page 485 and 486: 9.3. Linear Independence 471 Soluti
Page 487 and 488: 9.3. Linear Independence 473 Exerci
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Page 511 and 512: 9.6. Linear Transformations 497 The
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Page 515 and 516: 9.7. Isomorphisms 501 Example 9.66:
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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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A First Course in Linear Algebra, 2017a

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