Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A matrix is diagonalizable when algebraic and geometric multiplicities are equal

  1. Jul 21, 2009 #1
    A matrix is diagonalizable when algebraic and geometric multiplicities are equal.
    My professor proved this in class today, but I did not fully understand his explanation and proof. Can someone please help?
     
  2. jcsd
  3. Jul 22, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I thought I had answered this before. The "algebraic multiplicity" is the number of eigenvalues, counting "multiplicity". If follows that, including complex eigenvalues, since every polynomial can be factored into linear factors over the complex numbers, that the "algebraic multiplicity" of an n by n matrix is n. The "geometric multiplicity" is the number of independent eigenvalues. Thus, if the algebraic multiplicity is the same as the geometric multiplicity, the geometric multiplicity is also n and there exist n independent eigenvectors. But the "underlying" vector space of an n by n matrix has dimension n so those eigenvectors form a basis for that vector space. Writing the linear transformation this matrix represents in that vector space gives a diagonal matrix.
     
  4. Jul 23, 2009 #3
    Jack, find the eigenvalues and eigenvectors of A by hand(!)

    [tex]A = \begin{pmatrix}2 &1 &0\\0 &2 &0\\0 &0 &2\end{pmatrix}[/tex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: A matrix is diagonalizable when algebraic and geometric multiplicities are equal
  1. Diagonalizable matrix (Replies: 19)

Loading...