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Unformatted text preview: 18.06 Problem Set 4 Due Wednesday, 12 March 2008 at 4 pm in 2106. Problem 1: Do problem 2 from section 3.5 (pg. 168) in the book. Solution (10 points) We can test linear independence of vectors by putting them into the columns of a matrix A and finding the pivot columns. So, we define A = 1 1 1 1 1 1 1 1 1 1 1 1 and reduce: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 0 0 1 1 1 0 0 1 0 1 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 1 0 1 1 0 0 0 0 0 0 So, the matrix has rank 3, and thus at most three of the vectors can be linearly independent (for example the first three). Problem 2: Do problem 17 from section 3.5 (pg. 169). Solution (2+2+3+3 points) a) Any vector whose components are equal can be written v = a a a a = a 1 1 1 1 1 That is, V is just all scalar multiples of the vector (1 , 1 , 1 , 1), which means that the vector (1 , 1 , 1 , 1) is a basis. b) The vectors ( x 1 ,x 2 ,x 3 ,x 4 ) with components that add to zero are exactly the same as vectors in the nullspace of A = 1 1 1 1 We find a basis for a nullspace by taking all of the special solutions. In this case we end up with the vectors ( 1 , 1 , , 0), ( 1 , , 1 , 0) and ( 1 , , , 1). Of course, there are many other bases, which can be found just by looking at the space. c) We know that V is the orthogonal complement of the subspace spanned by the vectors (1 , 1 , , 0) and (1 , , 1 , 1). In this situation we can use the fact that the nullspace and row space are orthogonal complements. If we define the matrix A = 1 1 0 0 1 0 1 1 then V will be the nullspace of A . We find the basis for the nullspace as usual: A reduces to U = 1 1 0 0 1 1 1 which gives us special solutions ( 1 , 1 , 1 , 0) and ( 1 , 1 , , 1). These form a basis for V ....
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This note was uploaded on 08/10/2008 for the course MATH 250 taught by Professor Chanillo during the Spring '08 term at Rutgers.
 Spring '08
 CHANILLO
 Vectors, Linear Independence

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