lecture_34 - MA 265 LECTURE NOTES MONDAY APRIL 7 Orthogonal...

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Unformatted text preview: MA 265 LECTURE NOTES: MONDAY, APRIL 7 Orthogonal Complements Example. Consider the 3 × 3 matrix A =     1- 2 1 0 2 1- 1 4 1 3- 1 3 2 1- 1 2- 3 5 1 5     . We compute (1) the orthogonal complement for the row space of A and (2) the orthogonal complement for the column space of A . For (1), recall that this is simply the null space of A . We must take care though – The row space of A is a subspace of R 5 , whereas the null space of A will be a subspace of R 5 . Hence we transpose the solution at the end to make sure we are in the correct space. The equation A x = can best be solved using the reduced row echelon form for A : B =     1 0 7 2 4 0 1 3 1 1 0 0 0 0 0 0 0 0 0 0     = ⇒ x 1 + 7 x 3 + 2 x 4 + 4 x 5 = 0 x 2 + 3 x 3 + x 4 + x 5 = 0 Solving the system B x = , we see that x 3 = r , x 4 = s , and x 5 = t are arbitrary variables, so that       x 1 x 2 x 3 x 4 x 5       =      - 7 r- 2 s- 4 t- 3 r- s- t r s t       = r      - 7- 3 1       + s      - 2- 1 1       + t      - 4- 1 1       ....
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This note was uploaded on 03/11/2010 for the course MA 261A 0026100 taught by Professor ... during the Spring '10 term at Purdue University Calumet.

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lecture_34 - MA 265 LECTURE NOTES MONDAY APRIL 7 Orthogonal...

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