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!

Homework Help: Diagonalizability of matrix

  1. Jun 5, 2010 #1
    1. The problem statement, all variables and given/known data
    Suppose the A [tex]\in[/tex] Mn X n(F) has two distinct eigenvalues, [tex]\lambda[/tex]1 and [tex]\lambda[/tex]2, and that dim(E[tex]\lambda[/tex]1) = n -1. Prove A is diagonalizable.

    2. Relevant equations

    3. The attempt at a solution

    1. The charac poly clearly splits because we have eigenvalues.
    2. need to show m = dim (E).

    Ok, we are given that dim(E[tex]\lambda[/tex]1) = n - 1

    we know multiplicity has to be 1 [tex]\leq[/tex] dim(E[tex]\lambda[/tex]1) [tex]\leq[/tex] m.

    so: 1 [tex]\leq[/tex] n - 1 [tex]\leq[/tex] m.

    But im stuck now, not sure how to show that m = dim(E[tex]\lambda[/tex])
    Last edited: Jun 5, 2010
  2. jcsd
  3. Jun 5, 2010 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Re: Diagonalizability

    We're told that [tex]dimE^{\lambda_1}[/tex] is [tex]n-1[/tex]. What is [tex]dimE^{\lambda_2}[/tex] then, and what can you say about [tex]E^{\lambda_1} \oplus E^{\lambda_2}[/tex]?
  4. Jun 5, 2010 #3
    Re: Diagonalizability

    [tex]dimE^{\lambda_2}[/tex] = 1 then right?

    [tex]E^{\lambda_1} \oplus E^{\lambda_2}[/tex] = V?
  5. Jun 5, 2010 #4
    Re: Diagonalizability

    A linear operator T on a finite-dimensional vector space V is diagonalizable iff V is the direct sum of eigenspaces of T.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook