Week2 - LECTURE 1 I Inverse matrices We return now to the...

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Unformatted text preview: LECTURE 1 I. Inverse matrices We return now to the problem of solving linear equations. Recall that we are trying to find ~x such that A~x = ~ y. Recall: there is a matrix I such that I~x = ~x for all ~x ∈ R n . It follows that IA = A for all n × n matrices A . For the rest of the day, suppose A is an n × n matrix. Definition: Let A be an n × n matrix. We say a matrix B is an inverse for A if AB = BA = I . Notation: If B is an inverse for A , we write B = A- 1 , and we say A is invertible. Note that it is also true that B- 1 = A (check!) Theorem (Invertibility Theorem I): Suppose A has an inverse A- 1 . Then the equation A~x = ~ y has a unique solution, namely ~x = A- 1 ~ y. Remark: The converse of this theorem is also true, but we can’t prove it yet. We’ll do it later in the week when we discuss linear transformations. Proof. We have A~x = ~ y from which it follows that A- 1 A~x = A- 1 ~ y, and by definition of inverse, I~x = A- 1 ~ y, 1 and by definition of the identity, ~x = A- 1 ~ y. So if there’s a solution, this is it. To verify that this is a solution, check A ( A- 1 ~ y ) = ( AA- 1 ) ~ y = I~ y = ~ y which was the desired result. Theorem: Suppose B is inverse to A . Then B is the only inverse to A . Proof. Suppose B and B were two different inverses to A . Then we would have BAB = ( BA ) B = IB = B but on the other hand BAB = B ( AB ) = BI = B so B = B . Theorem: Suppose A and B are both invertible. Then AB is invertible, and its inverse is B- 1 A- 1 . Proof. We observe that ( AB )( B- 1 A- 1 ) = A ( BB- 1 ) A- 1 = AIA- 1 = AA- 1 = I . Left as exercise to check the other direction. Not all matrices have inverses. For instance Example: A = 1 0 0 0 does not have an inverse. Because the equation A~x = ~ has solutions y for any y ; in particular, the solution is not unique, so if A were invertible it would violate the theorem above. Some large classes of matrices have inverses. Example: Let D be a diagonal matrix, i. e. one which has all its nonzero entries on the diagonal. Suppose that D has no zeroes on the diagonal. Then D is invertible. 2 In fact, its inverse can be written in a very simple form. For example: 2 0 0 0 5 0 0 0 4 - 1 = 1 / 2 1 / 5 1 / 4 . Exercise: check this fact. Example: Let A = 1 1 1- 3 1 1 2- 8 . I claim that A has an inverse: namely, A- 1 = - 1 / 7 5 / 14 1 / 14 13 / 14- 9 / 28 1 / 28 3 / 14- 1 / 28- 3 / 28 To verify this, one needs only check that AA- 1 = A- 1 A = I . But how did I find the inverse? II. Gaussian elimination and the inverse In order to get this down, we begin with a definition. Definition: An elementary matrix is an n × n matrix with 1’s on the diagonal and exactly one non-zero entry off the diagonal....
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Week2 - LECTURE 1 I Inverse matrices We return now to the...

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