A First Course in Linear Algebra, 2017a

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170 R n 4.7.3 Projections In some applications, we wish to write a vector as a sum of two related vectors. Through the concept of projections, we can find these two vectors. First, we explore an important theorem. The result of this theorem will provide our definition of a vector projection. Theorem 4.37: Vector Projections Let⃗v and ⃗u be nonzero vectors. Then there exist unique vectors⃗v || and⃗v ⊥ such that where⃗v || is a scalar multiple of ⃗u, and⃗v ⊥ is perpendicular to ⃗u. ⃗v =⃗v || +⃗v ⊥ (4.13) Proof. Suppose 4.13 holds and ⃗v || = k⃗u. Taking the dot product of both sides of 4.13 with ⃗u and using ⃗v ⊥ •⃗u = 0, this yields ⃗v •⃗u =(⃗v || +⃗v ⊥ ) •⃗u = k⃗u •⃗u +⃗v ⊥ •⃗u = k‖⃗u‖ 2 which requires k = ⃗v •⃗u/‖⃗u‖ 2 . Thus there can be no more than one vector ⃗v || .Itfollows⃗v ⊥ must equal ⃗v −⃗v || . This verifies there can be no more than one choice for both⃗v || and⃗v ⊥ and proves their uniqueness. Now let ⃗v •⃗u ⃗v || = ‖⃗u‖ 2⃗u and let ⃗v •⃗u ⃗v ⊥ =⃗v −⃗v || =⃗v − ‖⃗u‖ 2⃗u Then⃗v || = k⃗u where k = ⃗v•⃗u . It only remains to verify⃗v ‖⃗u‖ 2 ⊥ •⃗u = 0. But ⃗v ⊥ •⃗u ⃗v •⃗u = ⃗v •⃗u − ‖⃗u‖ 2⃗u •⃗u = ⃗v •⃗u −⃗v •⃗u = 0 ♠ The vector⃗v || in Theorem 4.37 is called the projection of⃗v onto ⃗u and is denoted by ⃗v || = proj ⃗u (⃗v) We now make a formal definition of the vector projection.
4.7. The Dot Product 171 Definition 4.38: Vector Projection Let ⃗u and⃗v be vectors. Then, the (orthogonal) projection of⃗v onto ⃗u is given by ( ) ⃗v •⃗u ⃗v •⃗u proj ⃗u (⃗v)= ⃗u = ⃗u •⃗u ‖⃗u‖ 2⃗u Consider the following example of a projection. Example 4.39: Find the Projection of One Vector Onto Another Find proj ⃗u (⃗v) if ⎡ ⃗u = ⎣ 2 3 −4 ⎤ ⎡ ⎦,⃗v = ⎣ 1 −2 1 ⎤ ⎦ Solution. We can use the formula provided in Definition 4.38 to find proj ⃗u (⃗v). First, compute ⃗v •⃗u. This is given by ⎡ ⎤ 1 ⎡ ⎤ 2 ⎣ −2 ⎦ • ⎣ 3 ⎦ = (2)(1)+(3)(−2)+(−4)(1) 1 −4 = 2 − 6 − 4 = −8 Similarly, ⃗u •⃗u is given by ⎡ ⎣ 2 3 −4 ⎤ ⎡ ⎦ • ⎣ 2 3 −4 ⎤ ⎦ = (2)(2)+(3)(3)+(−4)(−4) = 4 + 9 + 16 = 29 Therefore, the projection is equal to ⎡ proj ⃗u (⃗v) = − 8 ⎣ 29 ⎡ − 16 29 = ⎢ − ⎣ 24 29 32 29 2 3 −4 ⎤ ⎥ ⎦ ⎤ ⎦ ♠
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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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Page 158 and 159: 144 R n Hence, every element in R 2
Page 160 and 161: 146 R n components are equal. Preci
Page 162 and 163: 148 R n Theorem 4.6: Properties of
Page 164 and 165: 150 R n 4.3 Geometric Meaning of Ve
Page 166 and 167: 152 R n 4.4 Length of a Vector Outc
Page 168 and 169: 154 R n Solution. We will use the f
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Page 172 and 173: 158 R n ⃗u 2⃗v ⃗u + 2⃗v ⃗
Page 174 and 175: 160 R n ⎡ Let ⃗q = ⎣ Then, x
Page 176 and 177: 162 R n Let t = x−2 y−1 3 ,t =
Page 178 and 179: 164 R n Example 4.26: Compute a Dot
Page 180 and 181: 166 R n Notice that this proof was
Page 182 and 183: 168 R n Therefore, the cosine of th
Page 186 and 187: 172 R n We will conclude this secti
Page 188 and 189: 174 R n Exercise 4.7.11 Find proj
Page 190 and 191: 176 R n Solution. By Definition 4.4
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Page 194 and 195: 180 R n Definition 4.48: Geometric
Page 196 and 197: 182 R n We will now look at an exam
Page 198 and 199: 184 R n 4.9.1 The Box Product Recal
Page 200 and 201: 186 R n = a ∣ e h ⎡ = det⎣ f
Page 202 and 203: 188 R n 4.10 Spanning, Linear Indep
Page 204 and 205: 190 R n 4.10.2 Linearly Independent
Page 206 and 207: 192 R n Theorem 4.66: Linear Indepe
Page 208 and 209: 194 R n Therefore we can write ⎡
Page 210 and 211: 196 R n Now suppose that⃗u ∉ sp
Page 212 and 213: 198 R n A subspace is simply a set
Page 214 and 215: 200 R n Definition 4.80: Basis of a
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Page 220 and 221: 206 R n Solution. An easy way to do
Page 222 and 223: 208 R n Since {⃗r 1 ,..., p⃗r j
Page 224 and 225: 210 R n Example 4.101: Rank, Column
Page 226 and 227: 212 R n Definition 4.106: Image of
Page 228 and 229: 214 R n Example 4.110: Null Space o
Page 230 and 231: 216 R n Theorem 4.114: Let A be an
Page 232 and 233: 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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