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11-coordinates

# 11-coordinates - COORDINATES HOMEWORK Section 3.4...

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COORDINATES Math 21b, O. Knill HOMEWORK: Section 3.4: 2,14,16,22,26,32*,38* B -COORDINATES. Given a basis ~v 1 , . . .~v n , define the matrix S = | . . . | ~v 1 . . . ~v n | . . . | . It is invertible. If ~x = i c i ~v i , then c i are called the B -coordinates of ~v . We write [ ~x ] B = c 1 . . . c n . If ~x = x 1 . . . x n , we have ~x = S ([ ~x ] B ). B -coordinates of ~x are obtained by applying S - 1 to the coordinates of the standard basis: [ ~x ] B = S - 1 ( ~x ) EXAMPLE. If ~v 1 = 1 2 and ~v 2 = 3 5 , then S = 1 3 2 5 . A vector ~v = 6 9 has the coordinates S - 1 ~v = - 5 3 2 - 1 6 9 = - 3 3 Indeed, as we can check, - 3 ~v 1 + 3 ~v 2 = ~v . EXAMPLE. Let V be the plane x + y - z = 1. Find a basis, in which every vector in the plane has the form a b 0 . SOLUTION. Find a basis, such that two vectors v 1 , v 2 are in the plane and such that a third vector v 3 is linearly independent to the first two. Since (1 , 0 , 1) , (0 , 1 , 1) are points in the plane and (0 , 0 , 0) is in the plane, we can choose ~v 1 = 1 0 1 ~v 2 = 0 1 1 and ~v 3 = 1 1 - 1 which is perpendicular to the plane. EXAMPLE. Find the coordinates of ~v = 2 3 with respect to the basis B = { ~v 1 = 1 0 , ~v 2 = 1 1 } . We have S = 1 1 0 1 and S - 1 = 1 - 1 0 1 . Therefore [ v ] B = S - 1 ~v = - 1 3 . Indeed - 1 ~v 1 + 3 ~v 2 = ~v .
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