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Matrix Representation of Complex Numbers

Date: 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


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)


     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   

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   
Associated Topics:
College Imaginary/Complex Numbers
College Linear Algebra

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