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Linear Algebra Diagonalizable Matrix

  1. Mar 9, 2013 #1
    1. The problem statement, all variables and given/known data
    For each of the following, give an example if it exists. If it doesn't exist, explain why.
    a) An invertible 3x3 matrix which is not diagonalizable.
    c) An 3x3 matrix A with A^6+I3=0 (Hint: Use the determinant)


    2. Relevant equations

    For a):

    I know in order for a matrix to be diagonalizable, the matrix A has to be similar with D (Diagonal Matrix). Which means having the same eigenvalues... Since the eigenvalues are on the main diagonal. All invertible matrix cant have 0 in the main diagonal in the reduced echelon form so it has to be diagonalizable so it does not exist...I am pretty sure this is not a good explanation, so I am asking for some clarification on diagonalization properties for invertible matrixe.

    for b): I used the determinant like this: DetA^6 = -detI3 => (DetA)^6= -1. So it does not exist...Is it what the question meant?

    I am not good with algebra so forgive me for my misunderstandings.


    3. The attempt at a solution
     
  2. jcsd
  3. Mar 9, 2013 #2

    Dick

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    For a) a 3x3 matrix is diagonalizable if it has three linearly independent eigenvectors. The eigenvalues don't have much to do with being diagonalizable. Do you know an example of a 2x2 matrix that doesn't have two linearly independent eigenvectors, hence is not diagonalizable? For b), that's exactly right.
     
  4. Mar 10, 2013 #3
    1 1
    0 1

    That is from an example that teacher gave us. But how do you know the eigenvectors without constructing examples and checking everytime? How to tell what the eigenvectors are by just the matrix itself?
     
  5. Mar 10, 2013 #4

    jbunniii

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    The eigenvalues of a triangular matrix are simply the diagonal entries, so 1 is the only eigenvalue of this matrix. So simply solve for the eigenvector(s):
    $$\begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} =\begin{bmatrix} x \\ y \end{bmatrix} $$
     
  6. Mar 10, 2013 #5
    How do you tell if it is diagonalizable or not? It is a 3x3 matrix so... I am not sure. I can find a invertible matrix, but how do you know if it is diagonalizable?
     
  7. Mar 10, 2013 #6

    vela

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    You generally have to calculate what the eigenvectors are.

    Go back to the 2x2 case you noted. You know from class it's not diagonalizable. You should verify this on your own because it'll help you figure out the 3x3 case. You should be able to convince yourself it's invertible as well. So in the 2x2 case, there is an invertible matrix which isn't diagonalizable. Think about how you might find a 3x3 matrix that's similar.
     
  8. Mar 10, 2013 #7
    can you just tell me how to find the diagonalizable matrix? Give me the basic steps just as a reminder. Thanks
     
  9. Mar 10, 2013 #8

    Dick

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    What about all of the hints you've been given don't you understand? You find eigenvectors by solving an algebra problem. It's easy to find a diagonizable matrix. You want to find one that's not diagonalizable. Work by analogy from the 2x2 case.
     
    Last edited: Mar 10, 2013
  10. Mar 11, 2013 #9
    1 1 1
    0 1 1
    0 0 1
     
  11. Mar 11, 2013 #10

    Dick

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    Yes, that works. Did you check that the only eigenvector is (1,0,0)?
     
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