A First Course in Linear Algebra, 2017a

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318 Linear Transformations system. By this, we mean that we replace ⃗ b by ⃗0 and look at A⃗x =⃗0. It turns out that there is a very important relationship between the solutions of the original system and the solutions of the associated homogeneous system. In the following theorem, we use linear transformations to denote a system of equations. Remember that T (⃗x)=A⃗x. Theorem 5.65: Particular Solution and General Solution Suppose⃗x p is a solution to the linear system given by , T (⃗x)=⃗b Then if⃗y is any other solution to T (⃗x)= ⃗ b,thereexists⃗x 0 ∈ ker(T ) such that ⃗y =⃗x p +⃗x 0 Hence, every solution to the linear system can be written as a sum of a particular solution,⃗x p ,anda solution⃗x 0 to the associated homogeneous system given by T (⃗x)=⃗0. Proof. Consider⃗y −⃗x p =⃗y +(−1)⃗x p .ThenT (⃗y −⃗x p )=T (⃗y) − T (⃗x p ).Since⃗y and⃗x p are both solutions to the system, it follows that T (⃗y)= ⃗ b and T (⃗x p )= ⃗ b. Hence, T (⃗y)−T (⃗x p )=⃗b−⃗b =⃗0. Let⃗x 0 =⃗y−⃗x p . Then, T (⃗x 0 )=⃗0so⃗x 0 is a solution to the associated homogeneous system and so is in ker(T ). ♠ Sometimes people remember the above theorem in the following form. The solutions to the system T (⃗x)=⃗b are given by⃗x p + ker(T ) where⃗x p is a particular solution to T (⃗x)=⃗b. For now, we have been speaking about the kernel or null space of a linear transformation T .However, we know that every linear transformation T is determined by some matrix A. Therefore, we can also speak about the null space of a matrix. Consider the following example. Example 5.66: The Null Space of a Matrix Let ⎡ A = ⎣ 1 2 3 0 2 1 1 2 4 5 7 2 Find null(A). Equivalently, find the solutions to the system of equations A⃗x =⃗0. Solution. We are asked to find ⎤ ⎦ { } ⃗x : A⃗x =⃗0 . In other words we want to solve the system, A⃗x =⃗0. Let
⃗x = ⎡ ⎢ ⎣ x y z w ⎤ ⎥ ⎦ . Then this amounts to solving ⎡ ⎣ 1 2 3 0 2 1 1 2 4 5 7 2 ⎡ ⎤ ⎦⎢ ⎣ 5.9. The General Solution of a Linear System 319 x y z w ⎤ ⎡ ⎥ ⎦ = ⎣ This is the linear system x + 2y + 3z = 0 2x + y + z + 2w = 0 4x + 5y + 7z + 2w = 0 To solve, set up the augmented matrix and row reduce to find the reduced row-echelon form. ⎡ ⎣ 1 2 3 0 0 2 1 1 2 0 4 5 7 2 0 ⎤ ⎦ →···→ ⎡ ⎢ ⎣ 0 0 0 1 0 − 1 3 0 1 ⎤ ⎦ 4 3 0 5 3 − 2 3 0 0 0 0 0 0 This yields x = 1 3 z− 4 3 w and y = 2 3 w− 5 3z. Since null(A) consists of the solutions to this system, it consists ⎤ ⎥ ⎦ vectors of the form, ⎡ ⎢ ⎣ 1 3 z − 4 3 w 2 3 w − 5 3 z z w ⎤ ⎡ ⎥ ⎦ = z ⎢ ⎣ 1 3 − 5 3 1 0 ⎤ ⎡ ⎥ ⎦ + w ⎢ ⎣ − 4 3 2 3 0 1 ⎤ ⎥ ⎦ ♠ Consider the following example. Example 5.67: A General Solution The general solution of a linear system of equations is the set of all possible solutions. Find the general solution to the linear system, ⎡ ⎤ ⎡ ⎤ x ⎡ ⎤ 1 2 3 0 ⎣ 2 1 1 2 ⎦⎢ y 9 ⎥ ⎣ z ⎦ = ⎣ 7 ⎦ 4 5 7 2 25 w ⎡ given that ⎢ ⎣ x y z w ⎤ ⎡ ⎥ ⎦ = ⎢ ⎣ 1 1 2 1 ⎤ ⎥ ⎦ is one solution.
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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
Page 282 and 283: 268 Linear Transformations Exercise
Page 284 and 285: 270 Linear Transformations In this
Page 286 and 287: 272 Linear Transformations ⎡ ⎣
Page 288 and 289: 274 Linear Transformations ⎡ T
Page 290 and 291: 276 Linear Transformations ⎡ (b)
Page 292 and 293: 278 Linear Transformations The nece
Page 294 and 295: 280 Linear Transformations Then, co
Page 296 and 297: 282 Linear Transformations More gen
Page 298 and 299: 284 Linear Transformations Solution
Page 302 and 303: 288 Linear Transformations Definiti
Page 304 and 305: 290 Linear Transformations This tel
Page 306 and 307: 292 Linear Transformations Example
Page 308 and 309: 294 Linear Transformations 2. T is
Page 310 and 311: 296 Linear Transformations Proposit
Page 312 and 313: 298 Linear Transformations Proof. S
Page 314 and 315: 300 Linear Transformations You can
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Page 320 and 321: 306 Linear Transformations Example
Page 322 and 323: 308 Linear Transformations Since {T
Page 324 and 325: 310 Linear Transformations Find a b
Page 326 and 327: 312 Linear Transformations Theorem
Page 328 and 329: 314 Linear Transformations and one
Page 330 and 331: 316 Linear Transformations Hence
Page 334 and 335: 320 Linear Transformations Solution
Page 338 and 339: 324 Complex Numbers Theorem 6.1: Pr
Page 340 and 341: 326 Complex Numbers Example 6.5: Co
Page 342 and 343: 328 Complex Numbers Definition 6.10
Page 344 and 345: 330 Complex Numbers Exercise 6.1.5
Page 346 and 347: 332 Complex Numbers Example 6.14: P
Page 348 and 349: 334 Complex Numbers This requires r
Page 350 and 351: 336 Complex Numbers Therefore, x 3
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Page 369 and 370: 7.2. Diagonalization 355 One eigenv
Page 371 and 372: 7.2. Diagonalization 357 In words,
Page 373 and 374: 7.2. Diagonalization 359 Conversely
Page 375 and 376: 7.2. Diagonalization 361 Theorem 7.
Page 377 and 378: 7.2. Diagonalization 363 7.2.3 Comp
Page 379 and 380: 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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