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lecture_37

# lecture_37 - MA 35100 LECTURE NOTES FRIDAY APRIL 23...

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MA 35100 LECTURE NOTES: FRIDAY, APRIL 23 “Inverting” Nonsquare Matrices Given the system A ~x = ~ b , let V = im( A ). In the previous lecture, we defined a least-squares solution ~x * R m as a system A ~x * = proj V ~ b . Unfortunately, A is not a square matrix, so we use the next fact to remedy this. Theorem. Let A be an n × m matrix. a. A T A is an m × m matrix. b. ker( A T ) = im( A ) . c. ker( A ) = ker( A T A ). d. If rank( A ) = m then A T A is invertible. (a) If A is an n × m matrix and B is an q × n matrix, then B A is an q × m matrix. If we let B = A T then q = m . (b) First we show that ker( A T ) im( A ) . Explicitly, write A = ~v 1 ~v 2 · · · ~v m so that A T ~ y = ~v T 1 ~v T 2 . . . ~v T m = ~v 1 · ~ y ~v 2 · ~ y . . . ~v m · ~ y . If ~ y ker( A T ) then ~ y is orthogonal to each of the columns ~v i . Now im( A ) = span( ~v 1 ,~v 2 , . . . ,~v m ), so ~ y im( A ) . Conversely, if ~ y im( A ) then A T ~ y = ~ 0 so that ~ y ker( A T ). Hence im( A ) ker( A T ).

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lecture_37 - MA 35100 LECTURE NOTES FRIDAY APRIL 23...

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