>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.
>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?
What's the difference between A transpose and A inverse? A specific example(s) would help. I'd love to test my skills.