Algebraic Multiplicity and Eigenspace

  1. 1. The problem statement, all variables and given/known data
    Find h in the matrix A such that the eigenspace for lambda=5 is two-dimensional.

    A= [5,-2,6,-1] [0,3,h,0] [0,0,5,4] [0,0,0,1]
    A-lambda*I(n) = [0,-2,6,-1] [0,-2,h,0] [0,0,0,4] [0,0,0,-4]


    2. Relevant equations



    3. The attempt at a solution
    I'm not really sure how to do this. Is there some relation between algebraic multiplicity and the eigenspace of a matrix that would help?

    I tried solving this by row operations to solve for the eigenvectors in the hope that I would be able to eliminate values of h that would have resulted in more or less than a 2 dimensional eigenspace, but it didn't work out. This is the matrix I got (keep in mind this is after applying the lambda value to the diagonals) A= [0,-2,6,-1] [0,0, h-6,1] [0,0,0,4] [0,0,0,0].

    I'm pretty confused about this any help would be greatly appreciated. Thanks.
     
  2. jcsd

  3. Hey there! the eigenspace of having lambda=5 is exactly the nullspace of A-lambda I . and since it is 2 dimensional , it suggest that the rank of the matrix is ?
     
  4. So the rank = 2 since rank = # columns (4 in this case) - dimNul A (in this case 2). So if the rank is to equal 2 then I will need another free variable, or I need to remove a pivot position. So since h-6 is in a pivot position I can easily make it a nonpivot column by setting h=6. This would ensure that the dimension of the null space is 2.

    Am I going the right way with this? It makes sense to me if this is right. Thanks for the help!
     
  5. HallsofIvy

    HallsofIvy 40,687
    Staff Emeritus
    Science Advisor

    So any eigenvector of A corresponding to eigenvalue 5 must satisfy
    [tex]\begin{bmatrix}0 & -2 & 6 & 1 \\ 0 & -2 & h & 0 \\ 0 & 0 & 0 & 4 \\ 0 & 0 & 0 & -4\end{bmatrix}\begin{bmatrix}w \\ x \\ y \\ z\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \\ 0\end{bmatrix}[/tex].

    That gives the equations -2x+ 6y+ z= 0, -2x+ hy= 0, 4z= 0, and -4z= 0. The last two obviously give z= 0 so the first two equations become -2x+ 6y= 0 and -2x+ hy= 0. One obvious eigenvector is (u, 0, 0, 0). There will be another, independent, eigenvector, and so the eigenspace will be two dimensional if and only if there exist non-zero x and y satifying both -2x+ 6y= 0 and -2x+ hy= 0.


     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?