A First Course in Linear Algebra, 2017a

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176 R n Solution. By Definition 4.42, P is a point in the plane if it satisfies the equation ⃗n • ( −→ 0P − −→ 0P 0 )=0 Given the above⃗n, P 0 ,andP, this equation becomes ⎡ ⎤ ⎛⎡ ⎤ ⎡ ⎤⎞ 1 5 2 ⎣ 2 ⎦ • ⎝⎣ 4 ⎦ − ⎣ 1 ⎦⎠ = 3 1 4 ⎡ ⎣ 1 2 3 ⎤ ⎛⎡ ⎦ • ⎝⎣ = 3 + 6 − 9 = 0 3 3 −3 ⎤⎞ ⎦⎠ Therefore P =(5,4,1) is contained in the plane. ⎡ Suppose⃗n = ⎣ Then a b c ⎤ ⎦, P =(x,y,z) and P 0 =(x 0 ,y 0 ,z 0 ). ⎡ ⎣ a b c ⎤ ⎛⎡ ⎦ • ⎝⎣ ⎡ ⎣ x y z a b c ⃗n • ( −→ 0P − −→ 0P 0 ) = 0 ⎤ ⎡ ⎤⎞ x 0 ⎦ − ⎣ y 0 ⎦⎠ = 0 z 0 ⎤ ⎡ ⎦ • ⎣ ⎤ x − x 0 y − y 0 ⎦ = 0 z − z 0 a(x − x 0 )+b(y − y 0 )+c(z − z 0 ) = 0 ♠ We can also write this equation as ax + by + cz = ax 0 + by 0 + cz 0 Notice that since P 0 is given, ax 0 + by 0 + cz 0 is a known scalar, which we can call d. This equation becomes ax + by + cz = d Definition 4.44: Scalar Equation of a Plane ⎡ ⎤ a Let ⃗n = ⎣ b ⎦ be the normal vector for a plane which contains the point P 0 =(x 0 ,y 0 ,z 0 ).Then if c P =(x,y,z) is an arbitrary point on the plane, the scalar equation of the plane is given by where a,b,c,d ∈ R and d = ax 0 + by 0 + cz 0 . ax + by + cz = d
4.8. Planes in R n 177 Consider the following equation. Example 4.45: Finding the Equation of a Plane ⎡ Find an equation of the plane containing P 0 =(3,−2,5) and orthogonal to⃗n = ⎣ −2 4 1 ⎤ ⎦. Solution. The above vector⃗n is the normal vector for this plane. Using Definition 4.42, we can determine the vector equation for this plane. ⃗n • ( −→ 0P − −→ 0P 0 ) = 0 ⎡ ⎣ −2 4 1 ⎤ ⎛⎡ ⎦ • ⎝⎣ ⎡ ⎣ x y z −2 4 1 ⎤ ⎡ ⎦ − ⎣ ⎤ ⎡ ⎦ • ⎣ 3 −2 5 x − 3 y + 2 z − 5 ⎤⎞ ⎦⎠ = 0 ⎤ ⎦ = 0 Using Definition 4.44, we can determine the scalar equation of the plane. −2x + 4y + 1z = −2(3)+4(−2)+1(5)=−9 Hence, the vector equation of the plane is ⎡ ⎤ ⎡ −2 ⎣ 4 ⎦ • ⎣ 1 x − 3 y + 2 z − 5 ⎤ ⎦ = 0 and the scalar equation is −2x + 4y + 1z = −9 ♠ Suppose a point P is not contained in a given plane. We are then interested in the shortest distance from that point P to the given plane. Consider the following example. Example 4.46: Shortest Distance From a Point to a Plane Find the shortest distance from the point P =(3,2,3) to the plane given by 2x + y + 2z = 2, and find the point Q on the plane that is closest to P. Solution. Pick an arbitrary point P 0 on the plane. Then, it follows that −→ −→ QP = proj ⃗n P0 P and ‖ −→ QP‖ is the shortest distance from P to the plane. Further, the vector −→ 0Q = −→ 0P− −→ QP gives the necessary point Q.
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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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Page 148 and 149: 134 Determinants The cofactor matri
Page 150 and 151: 136 Determinants Therefore, ∣ ∣
Page 152 and 153: 138 Determinants After row operatio
Page 154 and 155: 140 Determinants Using the given po
Page 156 and 157: 142 Determinants Does there exist a
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
Page 170 and 171: 156 R n = √ 1 + 9 + 16 = √ 26 I
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 184 and 185: 170 R n 4.7.3 Projections In some a
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 192 and 193: 178 R n From the above scalar equat
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 218 and 219: 204 R n ⎡ ⎢ ⎣ 1 2 1 1 3 0 1 3
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
Page 234 and 235: 220 R n Exercise 4.10.17 Are the fo
Page 236 and 237: 222 R n Thse vectors can’t possib
Page 238 and 239: 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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