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What does this mean?

  1. Jan 3, 2009 #1
    this is exam revision not homework so feal free to help

    let lamda be an eigenvalue of T, and let P be a polynomial with coefficients in F, define the linear mapping S=p(T) and show that p(lamda) is an eigenvalue of S

    i know that an eigenvalue of T is a element vnot=0 such that T(v)=lamda v for some lamda in the field, and that the scalar lamda here is the eigenvalue, however i dont understand this question at all

    so lamda is our eigenvalue meaning there must be a corresponding eigenvector v not equal to zero but whats p and what does this mapping show? please help
     
  2. jcsd
  3. Jan 3, 2009 #2

    statdad

    User Avatar
    Homework Helper

    First,
    If [tex]A[/tex] is the matrix and [tex]\lambda[/tex] the eigenvector, then [tex]A \vec x = \lambda \vec x [/tex] for some vector [tex]x[/tex]. Note these facts.

    [tex]
    \begin{align*}
    A^n \lambda & = A^{(n-1)} (A \vec x) \\
    & = A^{(n-2)} A (\lambda \vec x)\\
    & = \hdots \\
    & = \lambda^n \vec x
    \end{align*}
    [/tex]

    so [tex] \lambda^n [/tex] is an eigenvalue of [tex]A^n [/tex], with the same eigenvector

    Next, if [tex] c [/tex] is any constant, then

    [tex]
    (cA^n) \vec x = c\left(A^n \vec x\right) = c \lambda^n \vec x
    [/tex]

    so [tex] c\lambda^n [/tex] is an eigenvalue of [tex] cA^n [/tex]

    Finally, if [tex] a, b [/tex] are constants, and [tex] m, n [/tex] are integers, consider the
    two-term polynomial [tex] p(s) = as^m + bs^n[/tex]. The polynomial [tex]p(A) [tex] is

    [tex]
    p(A) = a A^m + bA^n
    [/tex]

    which is a matrix the same size as [tex] A [/tex]. The product [tex] p(A) \vec x [/tex] is

    [tex]
    \begin{align*}
    (a A^m + b A^n) \vec x & = (a A^m) \vec x + (b A^n) \vec x \\
    & = \left(a \lambda^m\right) \vec x + \left(b \lambda^n\right) \vec x\\
    & = \left(a \lambda^m + b \lambda^n \right) \, \vec x \\
    & = p(\lambda) \, \vec x
    \end{align*}
    [/tex]

    That case does not have a constant term in the polynomial. If you have
    [tex]
    p(s) = as^m + bs^n + d
    [/tex]

    where [tex] d [/tex] is a constant, the appropriate modification is

    [tex]
    p(A) = aA^m + bA^n + d I_n
    [/tex]

    where [tex] I_n [/tex] is the identity matrix the same size as [tex] A [/tex]. Again, it is easy to show that

    [tex]
    \begin{align*}
    p(A) \, \vec x & = \left(a A^m + b A^n + dI_n\right) \, \vec x\\
    & = \left(a \lambda^m + b \lambda^n + d\lambda\right) \, \vec x\\
    & = p(\lambda) \, \vec x
    \end{align*}
    [/tex]

    so again [tex] p(\lambda) [/tex] is an eigenvalue of [tex] p(A) [/tex].

    The case for a general polynomial requires a little more notation but the steps are the same.

    The idea: if [tex] A [/tex] is the matrix for a linear operator, so is [tex] p(A) [/tex] for
    any polynomial [tex] p [/tex], and the eigenvalues behave ``as we expect them to''.
     
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