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253hw1s

# 253hw1s - ma253 Fall 2010 Problem Set 1 Solutions 1 We work...

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ma253 , Fall 2010 — Problem Set 1 Solutions 1. We work with the linear space R 2 . Fix two numbers a and b and consider the function f : R 2 R given by f x 1 x 2 = ax 1 + bx 2 . Check that f is a linear function. We need to check that f distributes over sums and allows us to pull out scalars. Sums We start with two vectors x 1 x 2 and u 1 u 2 . Their sum is x 1 + u 1 x 2 + u 2 . Now f x 1 + u 1 x 2 + u 2 = a ( x 1 + u 1 ) + b ( u 2 + v 2 ) = ax 1 + au 1 + bx 2 + bu 2 , while f x 1 x 2 + f ( u 1 u 2 = ax 1 + bx 2 + au 1 + bu 2 . So the two computations give the same answer, which says that f distributes over sums. Scaling We start with x 1 x 2 and a scalar α ; scaling the vector gives αx 1 αx 2 . Now we just compute the two things that are supposed to be equal: f αx 1 αx 2 = aαx 1 + bαx 2 = α ( ax 1 + bx 2 ) , while αf x 1 x 2 = α ( ax 1 + bx 2 ) . They are equal, so we’ve shown that we can pull out scalars. So we’ve shown that f satisfies the two requirements to be a linear function, and we’re done. 2. Still in R 2 , look at the special vectors e 1 = 1 0 and e 2 = 0 1 . Explain how to get any vector from these two, using only adding and scaling operations. Any vector in R 2 looks like m n for some two numbers m and n . Now just notice that m n = m 1 0 + n 0 1 = m e 1 + n e 2 . Never Give an Iguana Viagra – headline, Reuters, January 25, 2007

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3. Suppose we know that f : R 2 R is a linear function, that f ( e 1 ) = a and that f ( e 2 ) = b . Show that for any vector v = x 1 x 2 we must have f ( v ) = ax 1 + bx 2 . We use problem 2 together with the information that f is linear. f x 1 x 2 = f ( x 1 e 1 + x 2 e 2 ) = f ( x 1 e 1 ) + f ( x 2 e 2 ) because we can distribute over sums = x 1 f ( e 1 ) + x 2 f ( e 2 ) because we can pull out scalars = x 1 a + x 2 b
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253hw1s - ma253 Fall 2010 Problem Set 1 Solutions 1 We work...

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