[Nutz: last part of last equation should have been "lambda2 iB".] At 09:54 AM 2/3/2010, Henry Baker wrote:

The general 2-D linear mapping over reals can be represented by a 2x2 matrix of reals:

[x1 y1]M = [x2 y2]

But this mapping can also be represented by 2 complex numbers A, B, which representation also has 4 real parameters:

(x+iy) = A*(x+iy)+B^2*(x-iy), or more succinctly,

z=Az+B^2z', where z' is the conjugate of z.

(We'll see the reason for using B^2 instead of B shortly.)

Now the matrix corresponding to (A,B) is symmetric iff A is real, so the eigenvalues of this mapping are real:

lambda1,2 = A+-|B|^2 = A+-BB'

We could go to the trouble of solving for the eigenvectors, but they are completely obvious (& obviously orthogonal): B, iB.

Consider B:

A(B)+B^2(B)' = AB+BBB' = AB+(BB')B = (A+BB')B = lambda1 B

Consider iB:

A(iB)+B^2(iB)' = A(iB)-BBiB' = A(iB)-BB'(iB) = (A-BB')iB = lambda2 B

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I'm sure that something along these lines can be done with quaternions; Cayley and/or Hamilton must have already done this, but it's difficult to decode their writings.

More to come.