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math-2011-note11

# math-2011-note11 - Theorem 53(Thm 11.2 p.243 Let A =(a1 a2...

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Theorem 53 (Thm 11.2, p.243) . Let A = ( a 1 , a 2 , ··· , a m ) . If a 1 , a 2 , ··· , a m are linearly independent if and only if det A negationslash = 0 . Proof. a) “only if”: If a 1 , a 2 , ··· , a m are linearly independent, then A is one-to-one and onto, hence has an inverse function. Call it A - 1 . Then A - 1 A = I . Hence det A negationslash = 0 . b) “if”: If a 1 , a 2 , ··· , a m are linearly dependent, then det A = 0 . Theorem 54 (Variant of Thm 11.8, p.248) . The following statements are equivalent (a) An m × m matrix A = ( a 1 , a 2 , ··· , a m ) is a one-to-one and onto linear function from R m to R m . (b) column vectors a 1 , a 2 , ··· , a m are linearly independent. (c) det A negationslash = 0 . (d) A is invertible. Proof. We only need to show that (c) implies (d). (c) implies that A is one-to-one and onto, hence that it has an inverse function. Call it A - 1 . Given y 1 and y 2 , let y 1 = A x 1 and y 2 = A x 2 . Then s y 1 + r y 2 = A ( s x 1 + r x 2 ) , and hence A - 1 ( s y 1 + r y 2 ) = s x 1 + r x 2 = sA - 1 ( y 1 )+ rA - 1 ( y 2 ) . Thus A - 1 is a linear function, which can be regarded as an m × m matrix. And AA - 1 = A - 1 A = I

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