1. Not finding help here? Sign up for a free 30min 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!

Eigenvalue and diagonalisation question

  1. Apr 3, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the eigenvalues and eigenvectors of the following matrix: M =
    1 1
    0 1
    Can this matrix be diagonalised?

    2. Relevant equations

    3. The attempt at a solution

    The characteristic equation is [itex](1 - \lambda)^{2} = 0[/itex] which gives [itex]\lambda = 1[/itex]. Substitute [itex]\lambda = 1[/itex] and eigenvector = {x,y} into the eigenvalue equation gives the two equations x+y = x and y = y. The first equation implies that y = 0. The second equation is redundant. So, x is free to assume any complex value. So, the eigenvalue is 1 and the eigenvector is {1,0}.

    I think everything I have done so far is fine. If it isn't, please point out.

    The problem starts with the second part: 'Can this matrix be diagonalised?' I know that to diagonalise a matrix is equivalent to changing the basis of the matrix and the eigenvectors. I am not quite sure I get this or the fact that the eigenvectors have to span the ??? to accomplish the diagonalisation.

    Thanks in advance for any help.
     
  2. jcsd
  3. Apr 3, 2012 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi failexam! :smile:
    changing the basis doesn't change the behaviour …

    the space spanned by the eigenvectors of the original matrix is one-dimensional (the line y = o)

    how many dimensions is the space spanned by the eigenvectors of a diagonal matrix? :wink:

    (alternatively, if it was diagonalised, what would those diagonal entries be?)
     
  4. Apr 3, 2012 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    An n by n matrix is "diagonalizable" if and only if it has n independent eigenvectors. If a matrix is NOT diagonalizable, it can be put in "Jordan Normal Form". And the matrix in this problem already is in Jordan Normal Form!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Eigenvalue and diagonalisation question
  1. Eigenvalues question. (Replies: 5)

  2. Eigenvalue questions (Replies: 2)

Loading...