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: Make the representative (diagonal)matrix of L

  1. Sep 19, 2012 #1
    1. The problem statement, all variables and given/known data

    V is a linear vector space of dimension n
    [itex]\phi = det(X I_n -A) [/itex] equals a product of first degree factors.
    [itex]Spec(L)={λ_1,...,λ_k}[/itex] is the set of eigenvalues

    show that: if L is diagonizable than d(λ_i)=m(λ_i)

    2. Relevant equations

    d_i=d(λ_i)= geometric multiplicity = dim E_i = dim {v in V| λ_i v = L(v) = A.v}
    m_i=m(λ_i)= algebraic multiplicity = multiplicity at which λ_i is a zero-point of [itex] \phi = det(X I_n - A) [/itex] . Because phi equals a product of first degree factors, I guess that [itex]\phi=(X-λ_i)^{m_i}[/itex]

    3. The attempt at a solution

    If L is diagonizable, then V has a basis β of eigenvectors of L. write the eigenvectors v so that they correspond to the eigenvalues they belong to. the first m_1 eigenvectors (note that m_1= m(λ_1) will correspond to the eigenvector λ_1. [itex]{ v_{1,1},v_{1,2},...,v_{1,m_1},v_{2,1},...,v_{2,m_2},...,v_{k,1},...,v_{k,m_k} } [/itex] is the set of eigenvectors. Since β is a basis of V, it should have n eigenvectors (thus [itex] m_1+m_2+...+m_k = n[/itex]).
    Now I am supposed to make the diagonal matrix of L with respect to β to show that d_i=m_i for all λ_i.
  2. jcsd
  3. Sep 19, 2012 #2


    User Avatar
    Science Advisor

    I don't understand why you would need to "make the diagonal matrix of L" to prove this. What you have is correct- there exist a basis of eigenvectors of L so there are a total of n independent eigenvectors. Since the number of eigenvectors corresponding to any one eigenvector cannot be larger than its algebraic multiplicity, in order to get all n eigenvectors you must have each eigenvalue must have number of eigenvectors (its geometric multiplicity) equal to its algebraic eigenvalue.
  4. Sep 19, 2012 #3
    So, first, we need n independent eigenvectors to form a basis for V. this means that the sum of d_i's will be n.
    I understand there is a theorem which says that [itex] d_i ≤ m_i[/itex], because phi will be of the form ( X - λ )^d * p(X). This means that there could be more zero-points of phi, than d.
    So we know [itex]Ʃd_i = n[/itex] and [itex] m_i ≤ d_i [/itex] for all i.
    since the Ʃm_i≤ n - the number of zeropoints will be smaller or equal to the total number of dimensions - we know that m_i=d_i for all i ?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook