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Homework week 14 1. Explain why the following set of vectors isNOT basis for the indicated vector spaces 1 2 u 0 3 u 2 7 in V 1 3 2 u 6 1 1 in V b) u Solutions: (a) A basis for has two linearly independent vectors therefore this set has one to many elements (b) A basis for has three independent vectors therefore the set is one element to short 2. Which of the following sets of vectors are basis for ? a) S 2 1 3 0 b) S 0 0 1 3 Solution: (a) yes: we can write a generic vector of in the unique form a b c 2 1 e 3 0 2c 3e c 0 a) u1 2 1 2 3 2 3 2 3 2 2 and the eqquation this defines is 2c which has solution if A 3e a 3e b 2 3 0 3 has an inverse det A 3 which implies that in general we can always find a set of coeficients c e such that the generic vector a b can be reconstructed in a unique way. This garanties also that 2c 3e 0 3e 0 is satified only by the trivial solution c 0 and e 0 therefore the set S is linearly independent and so the set of vectors forms a basis for 3 (b) No: We saw a theorem during lectures that states that if the zero is included in the set then automaticaly the set is linearly dependent. 1 1 0 0 1 0 0 0 0 is a basis for the vector space 0 0 0 0 1 0 0 1 M22 All 2 2 matrices by demonstrating that a) the set is linearly independent and b) that the these a b matrices define a generic element of M 22 , lets say the matrix .Solution: c d Taking the generic 2x2 matrix 3. Show that the set S a b c d a b 00 10 c 01 00 d 00 01 0 00 b0 0c 00 00 0c 0 b d 1 0 0 0 a 0 a c which demonstrates that the basis generates any matrix from M 22 . It only rest to prove that the set of matrices in S are linearly independent. To do that we write down the equation 0 0 0 0 k1 1 0 0 0 k1 k2 k3 k4 k2 0 1 0 0 k3 0 0 1 0 k4 0 0 0 1 which indicates that the only values of k s that make this equation true are k1 k2 k3 k4 0 which acording to the definition, implies that the set S is linearly independent. In conclusion the set S is a basis of M22 . 2