Invertible MatricesDate: 12/07/2001 at 09:17:44 From: Julie Subject: Invertible matrices Here is my problem: show that if A and B are nxn invertible matrices, then A^(-1) = (A+B)^(-1) + (A+AB^(-1)A)^(-1); assume that the expressions in the parenthesis on the right are invertible. I've tried substitution and multiplying by inverses (repeatedly). I don't know where to go from here. Can you help? Date: 12/08/2001 at 20:03:15 From: Doctor Pete Subject: Re: Invertible matrices Hi, and thanks for writing to Dr. Math. To make the work a little more compact, I will use the following notation for the inverse of a matrix: We'll write M' = M^(-1). Next, it might be useful to do a little preliminary work, which we might call a lemma: If C = AB, then what is C' = (AB)'? Well, we have B = A'AB = A'C, and A = ABB' = CB', so C = AB = (CB')(A'C) = CB'A'C or I = B'A'C, or C' = B'A'. Therefore, for any two nxn matrices A, B, we have [Lemma.] (AB)' = B'A'. Next, we want to find (A+B)'. We have (A+B)'(A+B) = (A+B)'A + (A+B)'B = I, so (A+B)'A = I - (A+B)'B. Hence [1] (A+B)' = (A+B)'AA' [2] = (I - (A+B)'B)A' [3] = A' - (A+B)'BA' [4] = A' - (A+B)'(AB')' [5] = A' - ((AB')(A+B))' [6] = A' - (AB'A + ABB')' [7] = A' - (AB'A + A)'. If we then add (AB'A + A)' to both sides, we find (AB'A + A)' + (A+B)' = A', which is equivalent to what was to be shown. Notice that I used the Lemma twice, once in step 4 and once in step 5. A little tricky but really quite elegant. - Doctor Pete, The Math Forum http://mathforum.org/dr.math/ |
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