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Find the bases for the eigenspaces of the matrix

  1. May 9, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the bases for the eigen spaces? I handwrote the question and my work so it would be easier to read and attached it so see attachment.


    2. Relevant equations



    3. The attempt at a solution
    My work is on the attachment. My question is this what do I do when I have 3 eigenvalues like in this case -1, 6, 6 and two are the same like how do I find for a third base it makes no sense to me?
     

    Attached Files:

  2. jcsd
  3. May 10, 2012 #2

    Mark44

    Staff: Mentor

    You have a mistake in your calculation for the eigenvectors for λ = 6. For this eigenvalue, there are two eigenvectors.

    From your equations,
    x1 = (3/4)x2
    x2 = .......x2
    x3 = ............x3

    Both x2 and x3 are arbitrary. You should be able to get two eigenvectors from the above.
     
  4. May 10, 2012 #3
    Thanks man I got it but can you tell me how abritary values are chosen I mean what if I set x1 to arbitrary?
     
  5. May 10, 2012 #4

    Mark44

    Staff: Mentor

    You can do that, but I prefer to use non-leading entries in each pivot row as the arbitrary values, together with any variables whose coefficients are zero.

    In the completely row-reduced matrix you end up with
    $$\begin{bmatrix} 1 & -3/4 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$

    This matrix represents this equation:
    x1 - (3/4)x2 = 0

    Since we have one equation but three variables, we're going to have two arbitrary variables.

    The leading entry (1) is the coefficient of x1. The non-leading entry is -3/4, so I choose x2 to be arbitrary. The coefficient of x3 is zero, so I choose it to be arbitrary, too.

    From the equation above, we have
    x1 = (3/4)x2
    x2 = x2 (arb.)
    x3 = x3 (arb.)

    If you line them up as I did in my previous post you can pick off the eigenvectors by sight.

    $$\begin{bmatrix} x_1 \\ x_2 \\ x_3\end{bmatrix} = x_2\begin{bmatrix} 3/4 \\ 1 \\ 0\end{bmatrix} + x_3\begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix}$$
     
  6. May 10, 2012 #5
    Thanks man that cleared it up for me
     
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