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