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hwk6

# hwk6 - Math 320 Fall 2010 HW#6 Selected Solutions Problem...

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Math 320 - Fall 2010 HW #6 Selected Solutions Problem 4.1.32 Show that the given set V is closed under addition and under multiplication by scalars and is therefore a subspace of R 3 . V is the set of all ( x, y, z ) such that z = 2 x + 3 y As instructed, to show V is a subspace of R 3 , it suffices to check that V is closed under addition and scalar multiplication. Let’s check addition first. Suppose v 1 and v 2 are two vectors in V (fancy notation: v 1 , v 2 V ). We need to check that v 1 + v 2 is in V . Let’s write v 1 = ( x 1 , y 1 , z 1 ) and v 2 = ( x 2 , y 2 , z 2 ), where the x 1 , x 2 , y 1 , y 2 , z 1 , z 2 are all arbitrary real numbers. Note that at this point, we haven’t yet used the definition of the subset V . Since v 1 and v 2 are vectors in V , we know that z 1 = 2 x 1 + 3 y 1 and z 2 = 2 x 2 + 3 y 2 . Now, let’s perform the addition: v 1 + v 2 = ( x 1 , y 1 , 2 x 1 + 3 y 1 ) + ( x 2 , y 2 , 2 x 2 + 3 y 2 ) = ( x 1 + x 2 , y 1 + y 2 , (2 x 1 + 3 y 1 ) + (2 x 2 + 3 y 2 )) How do we check that the sum v 1 + v 2 is in V ? Well, the ticket for admission into V is that z = 2 x + 3 y . In other words, to check that a vector is in V , we need to check that the z component is equal to twice the x component plus three times the y component. Let’s check: 2( x 1 + x 2 ) | {z } twice x component of v 1 + v 2 + 3( y 1 + y 2 ) | {z } three times y component of v 1 + v 2 = 2 x 1 + 2 x 2 + 3 y 1 + 3 y 2 = (2 x 1 + 3 y 1 ) + (2 x 2 + 3 y 2 ) | {z } z

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