Mid2Sol_4

Mid2Sol_4 - Since projL (v2 ) = we have 1 5 5 (v v1 )v1 =...

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Since proj L ( ~v 2 )= 1 k ~v 1 k 2 ( ~v 2 · ~v 1 ) ~v 1 = 5 3 ~v 1 = 5 3 ~v 1 +0 ~v 2 , we have [proj L ( ~v 2 )] { ~v 1 ,~v 2 } = µ 5 / 3 0 (b) Since proj L ( 1 ~v 1 , proj L ( ~v 2 5 3 ~v 1 , we conclude that the matrix of proj L in the basis ~v 1 ,~v 2 is [proj L ] { ~v 1 ,~v 2 } = µ 15 / 3 00 6. Let us Frst “orhonormalize” the vectors ~v 1 = 2 1 2 ,~ v 2 = 0 3 3 , i.e. Fnd an orthonormal basis of the plane. The Gram-Schmidt process gives ~u 1 = 2 / 3 1 / 3 2 / 3 u 2 = 2 / 3 2 / 3 1 / 3 . ±inally we use the formula: proj span( ~v 1 ,~v
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This note was uploaded on 04/12/2011 for the course MATH 33a taught by Professor Lee during the Fall '08 term at UCLA.

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