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Unformatted text preview: 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 )...
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