A First Course in Linear Algebra, 2017a

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420 Spectral Theory Example 7.89: Finding Eigenvalue and Eigenvector Find the eigenvalue and eigenvector for ⎡ which is closest to .9 + .9i. ⎣ 3 2 1 −2 0 −1 −2 −2 0 ⎤ ⎦ Solution. Form = ⎛⎡ ⎝⎣ ⎡ ⎣ 3 2 1 −2 0 −1 −2 −2 0 ⎤ ⎡ ⎦ − (.9 + .9i) ⎣ 1 0 0 0 1 0 0 0 1 ⎤⎞ ⎦⎠ −0.61919 − 10.545i −5.5249 − 4.9724i −0.37057 − 5.8213i 5.5249 + 4.9724i 5.2762 + 0.24862i 2.7624 + 2.4862i 0.74114 + 11.643i 5.5249 + 4.9724i 0.49252 + 6.9189i Then pick an initial guess an multiply by this matrix raised to a large power. ⎡ −0.61919 − 10.545i −5.5249 − 4.9724i ⎤ −0.37057 − 5.8213i ⎣ 5.5249 + 4.9724i 5.2762 + 0.24862i 2.7624 + 2.4862i ⎦ 0.74114 + 11.643i 5.5249 + 4.9724i 0.49252 + 6.9189i This equals ⎡ 1.5629 × 10 13 − 3.8993 × 10 12 i ⎤ ⎣ −5.8645 × 10 12 + 9.7642 × 10 12 i ⎦ −1.5629 × 10 13 + 3.8999 × 10 12 i Now divide by an entry to make the vector have reasonable size. This yields ⎡ ⎣ −0.99999 − 3.6140 × ⎤ 10−5 i 0.49999 − 0.49999i ⎦ 1.0 which is close to Then ⎡ ⎣ 3 2 1 −2 0 −1 −2 −2 0 ⎡ ⎣ ⎤⎡ ⎦⎣ −1 0.5 − 0.5i 1.0 −1 0.5 − 0.5i 1.0 ⎤ ⎦ ⎤ ⎡ ⎦ = ⎣ −1 −1.0 − 1.0i 1.0 1.0 + 1.0i Now to determine the eigenvalue, you could just take the ratio of corresponding entries. Pick the two corresponding entries which have the largest absolute values. In this case, you would get the eigenvalue is 1 + i which happens to be the exact eigenvalue. Thus an eigenvector and eigenvalue are ⎡ ⎣ −1 0.5 − 0.5i 1.0 ⎤ ⎦,1+ i ⎤ ⎦ 15 ⎡ ⎣ 1 1 1 ⎤ ⎦ ⎤ ⎦
7.4. Orthogonality 421 Usually it won’t work out so well but you can still find what is desired. Thus, once you have obtained approximate eigenvalues using the QR algorithm, you can find the eigenvalue more exactly along with an eigenvector associated with it by using the shifted inverse power method. ♠ 7.4.5 Quadratic Forms One of the applications of orthogonal diagonalization is that of quadratic forms and graphs of level curves of a quadratic form. This section has to do with rotation of axes so that with respect to the new axes, the graph of the level curve of a quadratic form is oriented parallel to the coordinate axes. This makes it much easier to understand. For example, we all know that x 2 1 + x2 2 = 1 represents the equation in two variables whose graph in R 2 is a circle of radius 1. But how do we know what the graph of the equation 5x 2 1 + 4x 1x 2 + 3x 2 2 = 1 represents? We first formally define what is meant by a quadratic form. In this section we will work with only real quadratic forms, which means that the coefficients will all be real numbers. Definition 7.90: Quadratic Form A quadratic form is a polynomial of degree two in n variables x 1 ,x 2 ,···,x n , written as a linear combination of x 2 i terms and x i x j terms. ⎡ ⎤ x 1 Consider the quadratic form q = a 11 x 2 1 + a 22x 2 2 + ···+ a nnx 2 n + a x 2 12x 1 x 2 + ···. We can write⃗x = ⎢ ⎥ ⎣ . ⎦ x n as the vector whose entries are the variables contained in the quadratic form. ⎡ ⎤ a 11 a 12 ··· a 1n a 21 a 22 ··· a 2n Similarly, let A = ⎢ ⎥ ⎣ . . . ⎦ be the matrix whose entries are the coefficients of x2 i and a n1 a n2 ··· a nn x i x j from q. Note that the matrix A is not unique, and we will consider this further in the example below. Using this matrix A, the quadratic form can be written as q =⃗x T A⃗x. q = ⃗x T A⃗x ⎡ = [ ] x 1 x 2 ··· x n ⎢ ⎣ ⎡ = [ ] x 1 x 2 ··· x n ⎢ ⎣ ⎡ ⎤ x 1 x 2 ⎢ ⎥ ⎣ . ⎦ x n ⎤ a 11 x 1 + a 21 x 2 + ···+ a n1 x n a 12 x 1 + a 22 x 2 + ···+ a n2 x n ⎥ . ⎦ a 1n x 1 + a 2n x 2 + ···+ a nn x n ⎤ a 11 a 12 ··· a 1n a 21 a 22 ··· a 2n ⎥ . . . ⎦ a n1 a n2 ··· a nn
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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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Page 463 and 464: Chapter 9 Vector Spaces 9.1 Algebra
Page 465 and 466: 9.1. Algebraic Considerations 451 E
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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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