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Linear algebra, basis, diagonal matrix

  1. Sep 4, 2011 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Write the A matrix and the x vector into a basis in which A is diagonal.
    [itex]A=\begin{pmatrix} 0&-i&0&0&0 \\ i&0&0&0&0 \\ 0&0&3&0&0 \\ 0&0&0&1&-i \\ 0&0&0&i&-1 \end{pmatrix}[/itex].
    [itex]x=\begin{pmatrix} 1 \\ a \\ i \\ b \\ -1 \end{pmatrix}[/itex].

    2. Relevant equations
    A=P^(-1)A'P.


    3. The attempt at a solution
    I found out the eigenvalues (spectra in fact) to be [itex]\sigma (A) = \{ -1,0,0,1,3 \}[/itex].
    I'm happy they told me A is diagonalizable; so I can avoid finding the Jordan form of A.
    So I know how is A'. I think that in order to find P, I must find the eigenvectors associated with each eigenvalues. P would be the matrix whose columns are the eigenvectors encountered. So with [itex]\lambda = -1[/itex], I found [itex]v_1=\begin{pmatrix} -i \\ i \\ 3 \\ 4-i \\ i+1 \end{pmatrix}[/itex]. But for [itex]\lambda =0[/itex] I reach the null vector as eigenvector, which I know is impossible.
    Am I doing something wrong?
     
  2. jcsd
  3. Sep 4, 2011 #2

    I like Serena

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    Hi fluidistic! :smile:

    What you're doing is right.
    But for lambda=0 you can still find a vector that is not the null vector.
    Actually there should be 2 linearly independent eigenvectors.
    They span the kernel of A.
     
  4. Sep 4, 2011 #3

    fluidistic

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    Ah true, I had made an arithmetic error. If there was no 2 l.i. eigenvectors for lambda=0 then the matrix A wouldn't be diagonalizable.

    The basis I found is [itex]P=B=\begin{pmatrix} 1&0&0&-i&0 \\ -i&0&0&i&0 \\ 0&0&0&3&-1 \\ 0&-i&-i&4-i&1 \\ 0&1&5&1+i& \frac{3}{5} \end{pmatrix}[/itex].
    What they ask for is [itex]A'=BAB^{-1}[/itex] which is (should be if I didn't make any error) just a diagonal matrix whose entries are the eigenvalues I found. And x'=Bx.
    Am I right?
     
  5. Sep 5, 2011 #4

    I like Serena

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    Didn't check for mistakes, but yes, you're right.
     
  6. Sep 5, 2011 #5

    lanedance

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    also not required, but couple of extra tips:

    - As A is close to diagonal, so you can shortcut and find the eigenvalues & eigenvectors for the 3 small blocks that make up A. These will be eigenvectors of A with zeroes in the other entries and will lead to simpler eigenvectors than the ones you have.

    - It always good to check by multiplying the matrix by the eigenvector you found and in fact I I think you need to do that. The eigenvectors you have don't look right at glance

    - If all your eigenvectors all turn out to be orthogonal, and you normalise, then finding the inverse of P is simple - its the transpose conjugate (P is unitary).

    - I think A is a normal matrix (A*A=AA*), if so it guarantees all eignespaces correponding to single eignevalues are orthogonal
     
    Last edited: Sep 5, 2011
  7. Sep 5, 2011 #6

    I like Serena

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    Note that A=A* (A is equal to its conjugate transpose).
    The so called spectral theorem tells us that this means that A is diagonizable.
     
  8. Sep 5, 2011 #7

    lanedance

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    So A is actually a hermitian matrix which guarantees real eigenvalues and diagonalisability

    hermitian matricies are also a subset of normal matricies, which gurantees diagonalisability by a unitary matrix
     
    Last edited: Sep 5, 2011
  9. Sep 5, 2011 #8

    vela

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    I'll second lanedance's recommendation to check your eigenvectors. You should be able to see by inspection that (0, 0, 1, 0, 0) is an eigenvector of A.

    I also found different eigenvalues for A, namely ±1, ±√2, and 3, so you should recheck that calculation first.
     
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