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

I Eigen Vectors, Geometric Multiplicities and more...

  1. Apr 10, 2016 #1
    My professor states that "A matrix is diagonalizable if and only if the sum of the geometric multiplicities of the eigen values equals the size of the matrix". I have to prove this and proofs are my biggest weakness; but, I understand that geometric multiplicites means the dimensions of the solution space for the equation Ax=λx (right?). But what does the "sum of the geometric multiplicities" mean? Could you point me in the right direction, thanks!!
     
    Last edited: Apr 10, 2016
  2. jcsd
  3. Apr 10, 2016 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    geometric multiplicity is the dimension of the solution space of ##\vec{\vec A}\vec x = \lambda_i \vec x## for one ##\lambda_i##. add them up for all ##i## and you get the sum of the geometric multiplicities, which you are asked to prove is equal to the size of A.
     
  4. Apr 10, 2016 #3
    How could I add up the dimensions? So for a 3x3 matrix that has three unique eigen vectors would I say that the dimension each of the eigen spaces is 3 and the sum of the geometric multiplicities is 3? Then would I say that A would have to be a square matrix of order 3?
     
  5. Apr 10, 2016 #4
    Is this in the right direction:
    In order for a matrix “A” to be diagonalizable then there is an equation such that P-1AP=D where D is the diagonalized matrix and P is the matrix formed from the Eigenvectors of A and if the sum of the geometric multiplicities is less than the size of A then P will not be invertible? Am I too far off, or did I assume something I shouldn't have?
     
  6. Apr 11, 2016 #5

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No, if an eigenvector ##\vec x## has a unique eigenvalue ##\lambda_x##, all multiples of that vector have the property ##
    \vec{\vec A}(p \vec x) = \lambda_x (p\vec x)\ ## (p a real number) so the dimension of the solution space is 1. Three unique eigenvalues let that add up to 3.

    If two vectors ##\vec x## and ##\vec y## have the same ##\lambda##, then ##p\vec x + q\vec y## has that too and the solution space for that degenerate eigenvalue has dimension 2. One other plus these 2 adds up to 3.
     
  7. Apr 11, 2016 #6
    => A is diagonalizable : ##A \sim \begin{pmatrix}\lambda_1 \text{ Id}_{m_1} & 0 & 0 \\
    0 & \ddots & 0\\
    0 & 0 & \lambda_p \text{ Id}_{m_p} \end{pmatrix}##. What is ##m_1,...,m_p## ? What is ##m_1 + ... + m_p ## equal to ?

    <= Say that matrix A represents an endomorphism on vector space ##E##. You are given that ## \text{dim}(E_{\lambda_1}) + ... + \text{dim}(E_{\lambda_p}) = \text{dim}(E) ##. Can you show that ##E=E_{\lambda_1} \bigoplus ... \bigoplus E_{\lambda_p} ## ? How does this prove that their exists a basis of ##E## in which A is diagonal ?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Eigen Vectors, Geometric Multiplicities and more...
Loading...