mat67-Homework3_Solutions

mat67-Homework3_Solutions - MAT067 University of California...

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MAT067 University of California, Davis Winter 2007 Solutions to Homework Set 3 1. Show that the vectors v 1 = (1 , 1 , 1), v 2 = (1 , 2 , 3), and v 3 = (2 , 1 , 1) are linearly independent in R 3 . Write the vector v = (1 , 2 , 5) as a linear combination of v 1 , v 2 , and v 3 . Solution : Let a 1 ,a 2 ,a 3 F , and suppose a 1 v 1 + a 2 v 2 + a 3 v 3 = ( x 1 ,x 2 ,x 3 ). Then a 1 + a 2 + 2 a 3 = x 1 a 1 + 2 a 2 a 3 = x 2 a 1 + 3 a 2 + a 3 = x 3 . (1) Subtracting the second equation from the first and third, 0 a 2 + 3 a 3 = x 1 x 2 a 1 + 2 a 2 a 3 = x 2 0 + a 2 + 2 a 3 = x 2 + x 3 . (2) Then, adding the third equation to first and subtracting 2 times the third equation from the second equation, 0 + 0 + 5 a 3 = x 1 2 x 2 + x 3 a 1 + 0 5 a 3 = 3 x 2 2 x 3 0 + a 2 + 2 a 3 = x 2 + x 3 . (3) Finally, upon dividing the first equation by 5, adding the first to the second equa- tion, and subtracting 2/5 times the first equation from the third equation, yields 0 + 0 + a 3 = 1 5 x 1 2 5 x 2 + 1 5 x 3 a 1 + 0 + 0 = x 1 + x 2 x 3 0 + a 2 + 0 = 2 5 x 1 1 5 x 2 + 3 5 x 3 . (4) Since setting ( x 1 ,x 2 ,x 3 ) = (0 , 0 , 0) yields a 1 = a 2 = a 3 = 0, the vectors v 1 , v 2 , and v 3 are linearly independent. On the other hand, setting ( x 1 ,x 2 ,x 3 ) = (1 , 2 , 5), we have a 1 = 6, a 2 = 3, and a 3 = 2. Hence, v = 6 v 1 + 3 v 2 + 2 v 3 .
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2 2. Consider the complex vector space V = C 3 and the list ( v 1 ,v 2 ,v 3 ) of vectors in V , where v 1 = ( i, 0 , 0), v 2 = ( i, 1 , 0), and v 3 = ( i,i, 1). Show that span( v 1 ,v 2 ,v 3 ) = V . Solution : Given an arbitrary vector v = ( a 1 ,a 2 ,a 3 ) C 3 , we need to show that there exist scalars c 1 ,c 2 ,c 3 C such that c 1 v 1 + c 2 v 2 + c 3 v 3 = v. For this to hold, we would need to have ( ic 1 + ic 2 + ic 3 ,c 2 + ic 3 , c 3 ) = ( a 1 ,a 2 ,a 3 ) , which implies that we need c 3 = a 3 , c 2 = a 2 ic 3 from which c 2 = a 2 + ia 3 , and c 1 = ia 1 c 2 c 3 from which c 1 = ia 1 a 2 + (1 i ) a 3 .
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3 3. Find a basis for the subspace U of R 5 defined by U = { ( x 1 ,x 2 ,...,x 5 ) | x 1 = 3 x 2 ,x 3 = 7 x 4 } . Solution : Proposition. Let U be the subspace of R 5 defined by U = { ( x 1 ,x 2 ,...,x 5 ) | x 1 = 3 x 2 ,x 3 = 7 x 4 } . Then a basis for U is given by the list S of vectors in R 5 , where S = ((3 , 1 , 0 , 0 , 0) , (0 , 0 , 7 , 1 , 0) , (0 , 0 , 0 , 0 , 1)) . Proof. Let U and S be as given. Then we need to check two things in order to prove that S is a basis for U . (a) First, we must show that U = span( S ). In other words, given any vector u U , we must show that there exist scalars a 1 ,a 2 ,a 3 R such that u = a 1 (3 , 1 , 0 , 0 , 0) + a 2 (0 , 0 , 7 , 1 , 0) + a 3 (0 , 0 , 0 , 0 , 1) Let u U . Then, to see that such scalars exist, note that u U implies that there are scalars x,y,z R such that u = (3 x,x, 7 y,y,z ). It follows that u = x (3 , 1 , 0 , 0 , 0) + y (0 , 0 , 7 , 1 , 0) + z (0 , 0 , 0 , 0 , 1) , and so u span( S ).
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