math54 - quiz 7 - solns - is -1-1 1 . 3. (4 pts) Suppose...

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Math 54 Quiz 7 Solutions 1. (3 pts) Find the matrix for the change of basis from B to C , where B = 1 - 1 0 , 0 1 0 , 0 0 1 , C = 0 1 0 , 1 - 1 0 , 0 1 1 The formula for the change of basis matrix is P C←B = 1 - 1 0 C 0 1 0 C 0 0 1 C which we can see from observation is P C←B = 0 1 - 1 1 0 0 0 0 1 (remember that [ ] C means the column vector of weights required to make the thing in parenthesis out of vectors of C ). 2. (3 pts) Find a basis for the nullspace and column space of the matrix A . A = 1 0 1 0 1 1 1 1 2 . The reduced row echelon form of A is 1 0 1 0 1 1 0 0 0 . The first two columns of this matrix are pivot columns, and therefore the first two columns of A form a basis of the columns space of A . Also, we can see that solution set to the homogeneous equation [ A | ~ 0] is x 1 = - x 3 , x 2 = - x 3 , x 3 = x 3
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from the reduced row echelon form. Therefore a basis of the null space
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Unformatted text preview: is -1-1 1 . 3. (4 pts) Suppose that V is a vector space with a basis B = { ~v 1 ,~v 2 } . Suppose that ~v V . State what vector space the vector [ ~v ] B is in. Suppose that [ ~v ] B is nonzero. Show that ~v would have to be nonzero as well. [ ~v ] B is, by denition, the column vector of weights required to make ~v as a linear combination of the vectors in B . Since B has two vectors in it, this column vector has two entries, so it is in R 2 . Suppose that [ ~v ] B = [ a b ]. Then, ~v = a~v 1 + b~v 2 . Since B is linearly independent, this linear combination can only be zero if a = b = 0. This was assumed not to be the case (since [ ~v ] B is nonzero), so ~v 6 = ~ 0....
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math54 - quiz 7 - solns - is -1-1 1 . 3. (4 pts) Suppose...

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