>>If you calculate det(<>-l*<>) for an n-by-n matrix <>, you >>will get a degree-n polynomial in l; generally you'll be expected to >>deal with 2-by-2 matrices, so this polynomial is just a quadratic. The >>solutions of this polynomial are your eigenvalues.
> Thanks, Grey Knight. I've got that part down.
>>The eigenvectors >>can then be found by solving the problem <> <x_i> = l_i <x_i>; that >>is (<> - l_i<>)<x_i> = <0>
> This is the part I don't get. I'm sorry but your notation loses me. I see the > "<>", and I think "dot product".
> We find the eigenvalues by solving matrix A....det (A-[lambda]*I ), where I is > the identity matrix.
What Gary is saying is this.... Suppose that A is a 2 by 2 matrix
(a b) (c d)
and then lambda is the eigenvalue. You should then solve for x1 and x2 in
(a-lambda b ) (x1) = (0) (c d-lambda) (x2) (0)
Although there are two equations here, one of them is redundant. So you can just solve either one... Since you have one equation, two unknowns, there can be infinite solutions on x1 and x2. For simplicity, just pick x1=1 and try to solve x2. If you can solve it, then you are done, after you normalize the eigenvectors to norm one. If you get contradiction by assuming x1=1, then x1 should be 0.
>>Two matrices are similar iff >> <> = <<X>> <> <<X'>>
> OK. So I = X*I*X'. So, X' is any matrix that when multiplied by X gives the > identity element as well?
Actually... I think there is a mistake here.
A n by n matrix X is similar to a n by n Y if you can find a n by n matrix T such that
X = T * Y * inv(T)
where inv(T) denotes the inverse of T, inv(T) * T = I
Inverse and transpose are two different concepts
If A is
1 2 3 4
then A' (A transpose) is
1 3 2 4
and A inverse is
( 4 -3 ) ( -2 1 ) / (-2)
> What's the difference between A transpose and A inverse? A specific example(s) > would help. I'd love to test my skills.