1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Eigenvalue of a matrix.

  1. Jan 14, 2012 #1
    1. The problem statement, all variables and given/known data

    αo, α1,..., αd [itex]\inℝ[/itex]. Show that αo + α1λ + α2λ2 + ... + αdλd [itex]\inℝ[/itex] is an eigenvalue of αoI + α1A + α2A2 + ... + αdAd [itex]\inℝ^{nxn}[/itex].


    2. The attempt at a solution

    If λ is an eigenvalue of A, then |A - Iλ| = 0. Also, λn is an eigenvalue An. So we basically have to somehow prove the following equation (after rearranging):

    1(A - Iλ) + α2(A2 - Iλ2) + ... + αd(Ad - Iλd)| = O

    3. Relevant equations

    I can't seem to get my head around this one. I almost used the triangular inequality to prove it before I realised that these are determinants we are dealing with, not absolute values. :/
     
  2. jcsd
  3. Jan 14, 2012 #2

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    So let [itex]P(z)[/itex] be a polynomial. We wish to prove that [itex]P(\lambda)[/itex] is an eigenvalue of [itex]P(A)[/itex].

    For each [itex]\mu\in \mathbb{C}[/itex], we can write

    [tex]P(z)-\mu=r_0(z-r_1)...(z-r_n)[/tex]

    Thus

    [tex]P(A)-\mu I = r_0(A-r_1 I)...(z-r_n I)[/tex]

    So [itex]P(A)-\mu I[/itex] is invertible iff all [itex](A-r_i I)[/itex] is invertible. What does that imply for the eigenvalues?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Eigenvalue of a matrix.
  1. Eigenvalues of a matrix (Replies: 22)

  2. Eigenvalue and matrix (Replies: 3)

Loading...