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By Waclaw Sierpinski, I. N. Sneddon, M. Stark

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F k :0 and A is invertible, so is kA and (kA)-' = k A-'. 2 The Special Case of "Square" Systems 7. 19 If A is invertible, so is A AT, and A*; moreover, (AT)-i = (A-' )T' (A-') _ (A)-', and (A*)-' = (A-')` 8. If A2=I", then A=A-'. 9. If A is invertible, so is AAT, AT A, AA*, and A*A. 10. If A and B are square matrices of the same size and the product AB is invertible, then A and B are both invertible. PROOF This is an exercise for the reader. There are many equivalent ways to say that a square matrix has an inverse.

Prove it is or provide a counterexample. 7. Cancellation laws: Suppose B and C are m-by-n matrices. Prove (i) if A is m-by-m and invertible and AB = AC, then B = C. (ii) if A is n-by-n and invertible and BA = CA, then B = C. 8. Suppose A and B are n-by-n and invertible. Argue that A-' + B-' _ A-'(A + B)B-'. If A and B were scalars, how would you have been led to this formula? (Hint: consider ( + h)). If (A + B)-' exists, what is (A-' + B-')-' in view of your result above? 9. Suppose U is n-by-n and U2 = I.

0 Another important lemma is given next. 2 Suppose A, B, (B + VA-1U), (A + UB-' V) are all invertible with Vs-by-n, A n-by-n, U n-by-s, and B s-by-s. Then (B + V A-' U)-' V A-' = B-' V (A + UB-'V)-1. PROOF Note VA-'(A+UB-'V)= V+VA-'UB-tV =(B+VA-'U) B-' V so V A-I (A + UB-' V) = (B + VA-' U)B-' V . Multiply both sides by (A + UB-' V)-' from the right to get VA-' = (B + VA-' U)B-' V (A + UB-'V)-'. Multiply both sides by (B + VA-1U)-1 from the left to get (B+VA-IU)-IVA-1 =B-IV(A+UB-'V)-1. 2 is given below.

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