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Unformatted text preview: Math 215 HW #4 Solutions 1. Problem 2.3.6. Choose three independent columns of U . Then make two other choice. Do the same for A . You have found bases for which spaces? U = 2 3 4 1 0 6 7 0 0 0 0 9 0 0 0 0 and A = 2 3 4 1 0 6 7 0 0 0 0 9 4 6 8 2 . Solution: The most obvious way to choose three independent columns of U is to pick the three pivot columns, namely 2 , 3 6 , and 1 9 . Another choice of three independent columns would be the first, third and fourth columns, which are independent because, if = c 1 2 + c 2 4 7 + c 3 1 9 = 2 c 1 + 4 c 2 + c 3 7 c 2 9 c 3 , then it must be the case that c 2 = c 3 = 0 (from the second and third components), so the first component reduces to 0 = 2 c 1 , so c 1 = 0 as well. We could also have chosen the second, third, and fourth columns. To see that these are independent, suppose = c 1 3 6 + c 2 4 7 + c 3 2 9 = 3 c 1 + 4 c 2 + c 3 6 c 1 + 7 c 2 9 c 3 . Then the third component implies that c 3 = 0 and the first two component give the system of equations 3 c 1 + 4 c 2 = 0 6 c 1 + 7 c 2 = 0 . Interpreting via the augmented matrix 3 4 6 7 and reducing by subtracting twice row 1 from row 2 yields 3 4 1 . 1 Then c 2 = 0 and, since 3 c 1 + 4 c 2 = 0, this means that c 1 = 0 as well. So, as indicated, the second, third, and fourth columns of U are also independent. Since U is just the echelon form of A (in particular, you can get from A to U by subtracting twice row 1 from row 4), the relationship between the columns of A is the same as the relationship between the columns of U . Hence, we can translate the three choices made above into three different choices of independent columns of A . Namely, the first, second, and fourth columns of A are independent, as are the first, third, and fourth columns of A , and the second, third, and fourth columns of A are also independent. 2. Problem 2.3.10. Find two independent vectors on the plane x + 2 y 3 z t = 0 in R 4 . Then find three independent vectors. Why not four? This plane is the nullspace of what matrix? Solution: Notice that 2 1 and 3 1 are vectors in the plane x + 2 y 3 z t = 0. They’re independent since = c 1 2 1 + c 2 3 1 = 2 c 1 + 3 c 2 c 1 c 2 implies that c 1 = 0 = c 2 . The vector 1 1 is also in this plane and we can see that this collection of three vectors is linearly independent as follows: suppose...
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 Spring '08
 ese302
 Linear Algebra, Vector Space

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