Eigenvalue/Eigenvector Proof for Matrix Polynomial p(A)

[cookie91]

Homework Statement

let p(x) = summation(from i=0 to k) aix^i

matrix polynomial for A is defined as p(A) = summation(i=0 up to k) aiA^i

Show that if (lambda, x) is an eigenpair of A then (p(lambda), x) is an eigenpair of p(A)

Homework Equations

The Attempt at a Solution

I pretty much have no idea where to start. I thought I could use Ax = lambda x like you would if you were proving that lambda^2 is an eigenvalue of A^2, etc, but I'm not sure how to get the p(A) bit?

 

[lanedance]

try multiplying the polynomial matrix form by the eignvector & see the results

 

[lanedance]

here's some latex to help, click on it to see code

$$\sum_{i=0}^k a_i x^i$$

 

[cookie91]

Thanks, I'm still not sure if I'm going about this the right way

so I've got

Use Ax = $$\lambda$$x

for p(A), multiply both sides of matrix polynomial by eigenvector x

p(A)x =
$$\sum_{i=0}^k a_i A^i$$ x

from the polynomial in the question, p(x) =
$$\sum_{i=0}^k a_i x^i$$

let x = $$\lambda$$

p($$\lambda$$) =
$$\sum_{i=0}^k a_i \lambda^i$$

p(A)x = $$\sum_{i=0}^k a_i \lambda^i$$ x

p(A)x = p($$\lambda$$x as required

but I'm not sure if I've really shown anything. how do i show that its the same x and $$\lambda$$ that belong to the matrix A?