Results 1 to 3 of 3
Like Tree3Thanks
  • 1 Post By emakarov
  • 2 Post By HallsofIvy

Math Help - what does an eigenvalue of 0 mean?

  1. #1
    Newbie
    Joined
    May 2013
    From
    smallville
    Posts
    6

    what does an eigenvalue of 0 mean?

    can a matrix be diagonalized if it has an eigenvalue of 0? I know that the diagonal must be invertible for this to happen but (yesterday I was solving a problem, a 3x3 matrix and I got 3 eigenvalues, 2 of which were 0, but if the two are 0, then the diagonal would look something like x 0 0/ 000/ 000 which I dont think is invertible... Am I going wrong somewhere? Also, if we were to find A^k=P^-1D^KP and we have such a matrix wouldnt we get an incorrect answer for A^K? I have a feeling I am missing something...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,517
    Thanks
    771

    Re: what does an eigenvalue of 0 mean?

    Recall that the column vector Ax is the linear combination of columns of A with coefficients from x. Therefore, if Ax = 0 for some x, then columns of A are linearly dependent, and so A is degenerate.
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,412
    Thanks
    1328

    Re: what does an eigenvalue of 0 mean?

    Well, I'm not certain what emakarov means by "degenerate". Certainly, if A has 0 as an eigenvalue, then there exist non-zero v such that Av= 0v= 0 so A is not invertible. But as far as "diagonalizable" is concerned, that depends entirely upon how many independent eigenvectors A has, not on what the eigenvalues are. An n by n matrix is diagonalizable if and only if it has n independent eigenvectors. For example, the matrix \begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix} has eigenvalue 1 and 0 and is diagonal while \begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix} has 0 as a double eigenvalue and is NOT Diagonalizable.

    (eigenvectors corresponding to distinct eigenvalues are necessarily independent so if an n by n matrix has n distinct eigenvalues, it has n independent eigenvectors and is diagonalizable. In order to have a non-diagonalizable matrix, we must have at least one eigenvalue with algebraic multiplicity greater than 1.)
    Thanks from topsquark and emakarov
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Eigenvalue
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 15th 2011, 06:39 PM
  2. eigenvalue help
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: January 25th 2010, 05:49 AM
  3. Eigenvalue
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: December 17th 2009, 08:22 PM
  4. Eigenvalue
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: April 12th 2009, 04:09 PM
  5. eigenvalue
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: October 16th 2008, 04:51 PM

Search Tags


/mathhelpforum @mathhelpforum