Lecture 24 - Mar 6

Lecture 24 - Mar 6 - 3/2 Then the matrix associated to T -1...

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Friday March 6 Lecture 24 : Properties of invertible matrices. (Refers to section 3.5) Expectations: 1. Recognize all characterizations of invertible matrices A . 24.1 Theorem For an n by n matrix A , the following are equivalent: 1) A is invertible 2) The rank of A is n . 3) A x = 0 only has the trivial solution. 4) A x = b has a unique solution for all vectors b . 5) A RREF = I . 6) Null( A ) = { 0 } 7) Col( A ) = R n . We outline the proofs in class. 24.2 Example – Suppose T is linear transformation mapping R 2 into R 2 rotating points counterclockwise by an angle of π /6. Find the matrix which rotates points clockwise by an angle of π /6. Solution: We first find the matrix A induced by T : T maps (1, 0) to ( 3/2, ½) and maps (0, 1) to (–1/2, 3/2) So the matrix A induced by T is
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3/2 –1/2, ½
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Unformatted text preview: 3/2 Then the matrix associated to T -1 is A-1 : 1/(3/4 + 1/4) 3/2 1/2, 3/2 24.3 Example If S and T are two 1-1 linear maps show that ( S T )-1 = T-1 S -1 . Solution. Let A and B be the two matrices induced by S and T respectively. Since both S and T are 1-1 then both A x = 0 and B x = 0 have unique solutions and so they are both invertible. (Previous result) Since A and B are both invertible then A-1 is induced by S -1 and B -1 is induced by T-1 . (Previous result) Furthermore AB is invertible and so ( AB ) -1 is induced by ( S T )-1 . Since T-1 S -1 ( x ) = B -1 A-1 ( x ) = ( AB )-1 ( x ) = ( S T )-1 ( x ) for all x . Then ( S T )-1 = T-1 S -1 ....
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Lecture 24 - Mar 6 - 3/2 Then the matrix associated to T -1...

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