the matrix A =
p1(x) = x^2 - 9, p2(x) = x + 3, p3(x) = x- 3. Show that p1(A)=p2(A)p3(A).
A = 2 0
I inserted the equation times the matrix A and then noticed p1(A) = (p2*p3)(A) and then decided to cancel A's. Which left only the equations and then it's obvious that by foiling p2*p3 you get p1. Way too easy. What am I missing?
they probably want you to actually compute A2 - 9I, and (A + 3I)(A - 3I):
but, yes, working backwards, we have, for any nxn matrix A:
it appears the exercise is designed to convince you that polynomials in a (square) matrix A work very similar to polynomials in x.
this lets us have polynomials "act" on vectors: given a vector v, and a polynomial p(x), and a matrix A, we define:
p(x).v = p(A)(v), so using a matrix A as a "go-between", we can multiply polynomial functions, and vectors, which is sort of weird, but it works.
polynomials and vector spaces have more in common than you might think.