Matrix Representation of Complex NumbersDate: 11/15/2000 at 16:19:40 From: John Jones Subject: Matrices I can't figure this math problem out: Find a matrix A and a matrix B such that (A+B) inverse = A inverse + B inverse Sorry, I don't have a key that makes it look like it (A+B) to the -1. Any help would be greatly appreciated. Thank you. Date: 11/15/2000 at 18:25:59 From: Doctor Schwa Subject: Re: Matrices John, Usually we write ^-1 to represent "to the -1." So, your question is to find matrices A and B such that (A+B)^-1 = A^-1 + B^-1 Multiplying both sides by (A+B), I get I = (A^-1 + B^-1)*(A + B) Distributing, I = I + (A^-1)*B + (B^-1)*A + I Hmm, that doesn't seem to help all that much. I wonder if there are real numbers a and b with a similar property, 1/(a+b) = 1/a + 1/b Multiplying both sides by ab gives ab / (a+b) = b + a and then multiplying by (a+b) gives ab = (a+b)^2 ab = a^2 + 2ab + b^2 which has some complex number solutions... Aha! There's a correspondence between complex numbers and matrices, namely that the number i acts the same as the matrix [0 -1] [1 0] You can check that: i^2 = [-1 0] [ 0 -1] So, if you put those pieces together, find some complex number solution to the equation at the top, and then translate the complex numbers into matrices, you'll have an answer. I wonder if there was some easier way that your teacher had in mind, though. I didn't see it... Feel free to write back if you have further questions (for example, if this partial solution makes no sense whatsoever). - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/ Date: 11/15/2000 at 19:32:23 From: John Jones Subject: Re: Matrices I partially understand what you mean, but I am having a hard time figuring out what you are doing with i. Date: 11/16/2000 at 15:34:20 From: Doctor Schwa Subject: Re: Matrices The question was how to find one example of matrices where (A+B)^-1 = A^-1 + B^-1. My method was to attempt an analogy, where a and b are real numbers, 1/(a+b) = 1/a + 1/b but it turns out that there aren't any real number solutions to this thing, so I resorted to complex numbers. You can set a equal to anything you want, and find a complex number b that solves the above equation (it's just a quadratic after you multiply away all the denominators). Now, suppose you have in hand complex numbers a and b that solve this equation. There's a matrix corresponding to every complex number. If b = x + iy, then the matrix B = [ x y] [-y x] will have the same behavior as the complex number b. Check it out: Try multiplying (1 + 2i)*(3 + 4i) and see what you get, then try multiplying [ 1 2] * [ 3 4] [-2 1] [-4 3] and you'll find that you get analogous answers. Similarly, the inverse matrix also works the same way as dividing by a complex number. So, once you have complex numbers a and b that solve your equation, you can translate them into the corresponding matrices and get an answer that will work. I still wonder if perhaps the teacher assigning this problem had a quite different method in mind. - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/