3.4+4.4

# 3.4+4.4 - Selected Answers to 3.4 and 4.4 Suggested...

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Selected Answers to 3.4 and 4.4 Suggested Problems 3.4 10. (a) S is clearly linearly independent, so we need only verify it is in V . 0 - 2+3 - 1+0 = 0, so it is. (b) We need a spanning set for V to which we can add S , then pare down into a basis. V has dimension 4, since determining any four entries ﬁxes the ﬁfth, but determining 3 leaves you freedom. Some vectors of V : (1 , 0 , 0 , 1 , 0), (0 , 1 , 0 , 0 , 1), (1 , 0 , - 1 , 0 , 1), (2 , 1 , 0 , 0 , 0). This is a linearly independent set and hence a basis. If we make a matrix with S as the ﬁrst column (this ensures S will be included in our basis) and this basis as the remaining four columns and row-reduce it, we get 1 0 0 - 1 0 0 1 0 1 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 . Since the ﬁrst, second, third, and ﬁfth columns correspond to the ﬁrst four elements of the standard basis, they correspond to our basis elements. Hence, a basis for V which includes S would be { (0 , 1 , 1 , 1 , 0) , (1 , 0 , 0 , 1 , 0) , (0 , 1 , 0 , 0 , 1) , (2 , 1 , 0 , 0 , 0)

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## This note was uploaded on 11/30/2009 for the course MATH 115A taught by Professor Liu during the Winter '07 term at UCLA.

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3.4+4.4 - Selected Answers to 3.4 and 4.4 Suggested...

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