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**Unformatted text preview: **Linear Algebra -115 Solutions to Eighth Homework Problem 9 (Section 2 . 4) Assume that AB is invertible. There is a proof using matrices that A,B are invertible, too, but it is rather tedious. We will ”translate” the problem to linear transformations and solve it quite easily: By Corollary 2 (page 102), A is invertible iff L A is. So, L AB = L A · L B is invertible, i.e. it is 1-1 and onto. By problem 12 of section 2 . 3 that we did last week, UT is 1-1 implies T is 1-1 and UT onto implies U is onto. In our case L A is onto and L B is 1-1. But A,B are square matrices, which means that L A ,L B are transformations from F n to F n . The dimensions of two spaces are equal, so once a transformation is one-to-one or onto it is an isomorphism. We see that L A ,L B are isomorphisms. Therefore A,B are invertible matrices. Now, if A = 1 1 = B t , then A,B are not invertible (recall here that only square matrices are invertible), but AB = (2) which is a 1 × 1 invertible matrix....

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