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Homework Help: Algebraic Multiplicity and Eigenspace

  1. Apr 9, 2010 #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. Apr 9, 2010 #2

    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. Apr 9, 2010 #3
    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. Apr 10, 2010 #4

    HallsofIvy

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    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.


     
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