A First Course in Linear Algebra, 2017a

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424 Spectral Theory = (U⃗y) T A(U⃗y) = ⃗y T (U T AU )⃗y = ⃗y T D⃗y The following procedure details the steps for the change of variables given in the above theorem. Procedure 7.93: Diagonalizing a Quadratic Form Let q be a quadratic form in the variables x 1 ,···,x n given by q = a 11 x 2 1 + a 22 x 2 2 + ···+ a nn x 2 n + a 12 x 1 x 2 + ··· Then, q can be written as q = d 11 y 2 1 + ···+ d nny 2 n as follows: 1. Write q =⃗x T A⃗x for a symmetric matrix A. 2. Orthogonally diagonalize A to be written as U T AU = D for an orthogonal matrix U and diagonal matrix D. ⎡ ⎤ y 1 y 2 3. Write⃗y = ⎢ . ⎣ .. ⎥. Then,⃗x = U⃗y. ⎦ y n 4. The quadratic form q will now be given by q = d 11 y 2 1 + ···+ d nny 2 n =⃗yT D⃗y where D = [ d ij ] is the diagonal matrix found by orthogonally diagonalizing A. Consider the following example. Example 7.94: Choosing New Axes to Simplify a Quadratic Form Consider the following level curve 6x 2 1 + 4x 1x 2 + 3x 2 2 = 7 shown in the following graph. x 2 x 1 Use a change of variables to choose new axes such that the ellipse is oriented parallel to the new coordinate axes. In other words, use a change of variables to rewrite q to eliminate the x 1 x 2 term.
7.4. Orthogonality 425 Solution. Notice that the level curve is given by q = 7forq = 6x 2 1 +4x 1x 2 +3x 2 2 . This is the same quadratic form that we examined earlier in Example 7.91. Therefore we know that we can write q = ⃗x T A⃗x for the matrix [ ] 6 2 A = 2 3 Now we want to orthogonally diagonalize A to write U T AU = D for an orthogonal matrix U and diagonal matrix D. The details are left to the reader, and you can verify that the resulting matrices are ⎡ U = D = ⎢ ⎣ 2√ 5 − 1 √ 5 ⎤ ⎥ 1√ √5 2 ⎦ 5 [ 7 0 0 2 [ ] y1 Next we write⃗y = .Itfollowsthat⃗x = U⃗y. y 2 We can now express the quadratic form q in terms of y, using the entries from D as coefficients as follows: ] q = d 11 y 2 1 + d 22 y 2 2 = 7y 2 1 + 2y2 2 Hence the level curve can be written 7y 2 1 + 2y2 2 = 7. The graph of this equation is given by: y 2 y 1 The change of variables results in new axes such that with respect to the new axes, the ellipse is oriented parallel to the coordinate axes. These are called the principal axes of the quadratic form. ♠ The following is another example of diagonalizing a quadratic form.
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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 437: 7.4. Orthogonality 423 We now turn
Page 441 and 442: 7.4. Orthogonality 427 = 2y 2 1 + 8
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Page 445 and 446: 7.4. Orthogonality 431 Exercise 7.4
Page 447 and 448: Chapter 8 Some Curvilinear Coordina
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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
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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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