A First Course in Linear Algebra, 2017a

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12 Systems of Equations 2. Next we want to prove that the systems 1.1 and 1.3 have the same solution set. That is E 1 = b 1 ,E 2 = b 2 has the same solution set as the system E 1 = b 1 ,kE 2 = kb 2 provided k ≠ 0. Let (x 1 ,···,x n ) be a solution of E 1 = b 1 ,E 2 = b 2 ,. We want to show that it is a solution to E 1 = b 1 ,kE 2 = kb 2 . Notice that the only difference between these two systems is that the second involves multiplying the equation, E 2 = b 2 by the scalar k. Recall that when you multiply both sides of an equation by the same number, the sides are still equal to each other. Hence if (x 1 ,···,x n ) is a solution to E 2 = b 2 , then it will also be a solution to kE 2 = kb 2 . Hence, (x 1 ,···,x n ) is also a solution to 1.3. Similarly, let (x 1 ,···,x n ) be a solution of E 1 = b 1 ,kE 2 = kb 2 . Then we can multiply the equation kE 2 = kb 2 by the scalar 1/k, which is possible only because we have required that k ≠ 0. Just as above, this action preserves equality and we obtain the equation E 2 = b 2 . Hence (x 1 ,···,x n ) is also a solution to E 1 = b 1 ,E 2 = b 2 . 3. Finally, we will prove that the systems 1.1 and 1.4 have the same solution set. We will show that any solution of E 1 = b 1 ,E 2 = b 2 is also a solution of 1.4. Then, we will show that any solution of 1.4 is also a solution of E 1 = b 1 ,E 2 = b 2 .Let(x 1 ,···,x n ) be a solution to E 1 = b 1 ,E 2 = b 2 .Then in particular it solves E 1 = b 1 . Hence, it solves the first equation in 1.4. Similarly, it also solves E 2 = b 2 . By our proof of 1.3,italsosolveskE 1 = kb 1 . Notice that if we add E 2 and kE 1 , this is equal to b 2 +kb 1 . Therefore, if (x 1 ,···,x n ) solves E 1 = b 1 ,E 2 = b 2 it must also solve E 2 +kE 1 = b 2 +kb 1 . Now suppose (x 1 ,···,x n ) solves the system E 1 = b 1 ,E 2 + kE 1 = b 2 + kb 1 . Then in particular it is a solution of E 1 = b 1 . Again by our proof of 1.3, it is also a solution to kE 1 = kb 1 . Now if we subtract these equal quantities from both sides of E 2 + kE 1 = b 2 + kb 1 we obtain E 2 = b 2 , which shows that the solution also satisfies E 1 = b 1 ,E 2 = b 2 . Stated simply, the above theorem shows that the elementary operations do not change the solution set of a system of equations. We will now look at an example of a system of three equations and three variables. Similarly to the previous examples, the goal is to find values for x,y,z such that each of the given equations are satisfied when these values are substituted in. Example 1.9: Solving a System of Equations with Elementary Operations Find the solutions to the system, ♠ x + 3y + 6z = 25 2x + 7y + 14z = 58 2y + 5z = 19 (1.5) Solution. We can relate this system to Theorem 1.8 above. In this case, we have E 1 = x + 3y + 6z, b 1 = 25 E 2 = 2x + 7y + 14z, b 2 = 58 E 3 = 2y + 5z, b 3 = 19 Theorem 1.8 claims that if we do elementary operations on this system, we will not change the solution set. Therefore, we can solve this system using the elementary operations given in Definition 1.6. First,
1.2. Systems of Equations, Algebraic Procedures 13 replace the second equation by (−2) times the first equation added to the second. This yields the system x + 3y + 6z = 25 y + 2z = 8 2y + 5z = 19 (1.6) Now, replace the third equation with (−2) times the second added to the third. This yields the system x + 3y + 6z = 25 y + 2z = 8 z = 3 (1.7) At this point, we can easily find the solution. Simply take z = 3 and substitute this back into the previous equation to solve for y, and similarly to solve for x. The second equation is now x + 3y + 6(3)=x + 3y + 18 = 25 y + 2(3)=y + 6 = 8 z = 3 y + 6 = 8 You can see from this equation that y = 2. Therefore, we can substitute this value into the first equation as follows: x + 3(2)+18 = 25 By simplifying this equation, we find that x = 1. Hence, the solution to this system is (x,y,z)=(1,2,3). This process is called back substitution. Alternatively, in 1.7 you could have continued as follows. Add (−2) times the third equation to the second and then add (−6) times the second to the first. This yields x + 3y = 7 y = 2 z = 3 Now add (−3) times the second to the first. This yields x = 1 y = 2 z = 3 a system which has the same solution set as the original system. This avoided back substitution and led to the same solution set. It is your decision which you prefer to use, as both methods lead to the correct solution, (x,y,z)=(1,2,3). ♠
Page 1: with Open Texts Linear A Algebra wi
Page 5: A First Course in Linear Algebra an
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Page 12 and 13: iv Table of Contents 3 Determinants
Page 14 and 15: vi Table of Contents 7.3.5 The Matr
Page 17 and 18: Chapter 1 Systems of Equations 1.1
Page 19 and 20: 1.1. Systems of Equations, Geometry
Page 21 and 22: 1.2. Systems of Equations, Algebrai
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Page 68 and 69: 54 Matrices Consider the following
Page 70 and 71: 56 Matrices Solution. Notice that b
Page 72 and 73: 58 Matrices Proposition 2.10: Prope
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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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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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A First Course in Linear Algebra, 2017a

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