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quiz2

# quiz2 - 0 0 0 f 0 0 0 0 0 0 0 0 1 so the set T = 1 0 0 0 0...

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Name: Linear Algebra Quiz 2 June 5, 2006 1. (10 points) Is the set { 1 2 - 1 , - 1 - 3 2 , 4 6 - 2 } linearly independent or dependent? Ex- plain. 2. (10 points) Consider the set S of all 3 × 3 upper triangular matrices, i.e., matrices of the form a b c 0 d e 0 0 f Show that S is a subspace of the set of all 3 × 3 matrices. Find a basis for S and calculate dim S . Solution: 1. Calculate 1 - 1 4 2 - 3 6 - 1 2 - 2 = 0 so that the set is linearly dependent. 2. Consider upper triangular matrices A = a 11 a 12 a 13 0 a 22 a 23 0 0 a 33 and B = b 11 b 12 b 13 0 b 22 b 23 0 0 b 33 Then A + B = a 11 + b 11 a 12 + b 12 a 13 + b 13 0 a 22 + b 22 a 23 + b 23 0 0 a 33 + b 33 which is upper triangular so that S is closed under addition. Let c be in R . Then cA = b 11 ca 12 ca 13 0 ca 22 ca 23 0 0 ca 33 which is again upper triangular so that S is closed under scalar multiplication. This shows that S is a subspace of the space of 3 × 3 matrices. Notice that any element in S may be written a b c 0 d e 0 0 f = a 1 0 0 0 0 0 0 0 0 + b 0 1 0 0 0 0 0 0 0 + c 0 0 1 0 0 0 0 0 0 + d 0 0 0 0 1 0 0 0 0 + e 0 0 0 0 0 1 0 0 0 + f 0 0 0 0 0 0 0 0 1 so the set T = { 1 0 0 0 0 0 0 0 0 , 0 1 0 0 0 0 0 0 0 , 0 0 1 0 0 0 0 0 0

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Unformatted text preview: 0 0 0 + f 0 0 0 0 0 0 0 0 1 so the set T = { 1 0 0 0 0 0 0 0 0 , 0 1 0 0 0 0 0 0 0 , 0 0 1 0 0 0 0 0 0 , 0 0 0 0 1 0 0 0 0 , 0 0 0 0 0 1 0 0 0 , 0 0 0 0 0 0 0 0 1 } spans S . T is certainly linearly independent since if we had a 1 0 0 0 0 0 0 0 0 + b 0 1 0 0 0 0 0 0 0 + 1 c 0 0 1 0 0 0 0 0 0 + d 0 0 0 0 1 0 0 0 0 + e 0 0 0 0 0 1 0 0 0 + f 0 0 0 0 0 0 0 0 1 = 0 0 0 0 0 0 0 0 0 then a b c d e 0 0 f = 0 0 0 0 0 0 0 0 0 so that a = b = c = d = e = f = 0. Hence the set T forms a basis for S . Since T has 6 elements, dim S = 6. 2...
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