Suppose I get the eigenvalues of A, which are

[sessomw5098]

Suppose I get the eigenvalues of A, which are $$\lambda_{1},\lambda_{2},\dots \lambda_{n}$$. Also, given any polynomial f(x), I get the eigenvalues of f(A). I'm trying to show that the eigenvalues of f(A) are $$f(\lambda_{1}),f(\lambda_{2}),\dots f(\lambda_{n})$$. Is this possible? How would I go about showing this?

 

[Deveno]

if f(x) = anx^n +...+a1x + a0, and A has an eigenvector v with eigenvalue λ,

then f(A)(v) = (anA^n +...+a1A + a0I)(v)

= an(A^n(v)) +...+ a1(A(v)) + a0(v) =

= an((λ^n)v) +...+ a1(λv) + a0(v)

= (anλ^n +...+a1λ + a0)(v) = f(λ)v, so f(λ) is an eigenvalue of f(A).

 

[tiny-tim]

hi sessomw5098!

(have a lambda: λ )

hint: if Ax = λx, what is Aⁿx ?