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Unformatted text preview: Math 235: Linear Algebra HW 4 Problem 2.3.2 Problem 2.3.2 Are the vectors 1 = (1 , 1 , 2 , 4) , 2 = (2 , 1 , 5 , 2) , 3 = (1 , 1 , 4 , 0) , and (2 , 1 , 1 , 6) linearly indepen dent in R 4 ? The vectors are not linearly independent since 1 + 2 3 4 = 0. Problem 2.3.3 Find a basis for the subspace of R 4 spanned by the vectors in 2.3.2. { 1 , 2 , 3 } are a basis of the subspace. Alternatively, you can construct a matrix whose rows are 1 , 2 , 3 and 4 , row reducing will give you a matrix with one row of zeros. The other three rows are a basis for the subspace. Additionally, this proves that the subspace has dimension 3, thus the three vectors are a basis. Problem 2.3.5 Find three vectors in R 3 which are linearly dependent, and such that any two of them are linearly independent. v 1 = (1 , 1 , 0), v 2 = (1 , 1 , 1) and v 3 = (0 , , 1) Problem 2.3.7 Let V be the vector space of all 2 X 2 matrices over a field F . Let W 1 be the set of matrices of the form x x y z and let W 2 be the set of matrices of the form a b a c (a) Prove that W 1 and W 2 are subspaces of V . Clearly the zero matrix is in both W 1 and W 2 , and since for any k F we have that k x x y z = kx kx ky kz W 1 and k a b a c = ka kb ka kc . We get negatives by letting k = 1. It is left to show that W 1 and W 2 are closed: Let x 1 x 1 y 1 z 1 W 1 and x 2 x 2 y 2 z 2 W 2 then x 1 x 1 y 1 z 1 + x 2 x 2 y 2 z 2 = x 1 + x 2 x 1 + x 2 y 1 + y 2 z 1 + z 2 W 1 . The same argument shows that W 2 is closed. (a) Problem 2.3.7 continued on next page... Page 1 of 6 Math 235: Linear Algebra HW 4 Problem 2.3.7 [(b)] (b) Find the dimension of W 1 , W 2 , W 1 + W 2 and W 1 T W 2 . A basis for W 1 is 1 1 , 0 0 1 0 , 0 0 0 1 , so W 1 has dimension 3. A basis for W 2 is 1 1 0 , 0 1 0 0 , 0 0 0 1 , so W 2 has dimension 3. Anything in W 1 T W 2 has the form a a a b . So W 1 T W 2 has a basis 1 1 1 , 0 0 0 1 , so it has dimension 2. We can now use theorem 6 on page 46 to find that dim( W 1 + W 2 ) = 4 (b) Problem 2.3.11 Let V be the set of real numbers. Regard V as a vector space over the field of rational numbers, with the usual operations. Prove that this vector space is not finitedimensional. Assume you have a finite basis { b 1 ,...,b n } let V be the vector space they generate. We will prove that V is too small to be R ....
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This note was uploaded on 11/14/2010 for the course PHYCS 498 taught by Professor Aa during the Spring '10 term at University of Illinois, Urbana Champaign.
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