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Eigenvalues of two matrices are equal

  1. Sep 18, 2013 #1
    Hi everyone,

    I have two matrices A and B,
    A=[0 0 1 0; 0 0 0 1; a b a b; c d c d] and B=[0 0 0 0; 0 0 0 0; 0 0 a b; 0 0 c d].
    I have to proves theoretically that two of the eigenvalues of A and B are equal and remaining two eigenvalues of A are 1,1.
    I tried it by calculating the determinant of A and B and I got close to the result but I am not able to prove it completely.

    I got result like this,
    sum of roots of determinant of A and B as
    p+q+r+s=p1+q1 (p,q,r,s are roots of det of A, p1,q1 are roots of det of B)

    Product of roots


    Please help me to show that two eigenvalues of A and B are equal.
  2. jcsd
  3. Sep 18, 2013 #2


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    Find the characteristic polynomials of A and B.

    Then factorize A - the question tells you two of the factors.
  4. Sep 18, 2013 #3
    I already tried that way.

    The characteristic equations that I got for A is
    p^4 - p^3(a+d) + p^2(ad-bc-a-d) +p (2ad-2bc) +ad-bc=0
    for B

    I cant factorize A polynomial equation, since it does not have simple 1 or -1 as roots.
  5. Sep 18, 2013 #4


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    The question says two roots of the A polynomial are equal to 1. So if (p-1)^2 isn't a factor, either you made a mistake somewhere, or the question is wrong.

    I agree with you that p-1 us not a factor of the A polynomial, so I think the question in your OP is wrong. Are you missing some minus signs in the A matrix?
  6. Sep 18, 2013 #5
    Matlab gives -1 as an eigenvalue but theoretically i cant prove it. There is no mistake in the theoretical proof, i checked it many times. The signs in A matrix are also correct.
  7. Sep 18, 2013 #6
    This is an example of A matrix that I have
    0 0 1 0
    0 0 0 1
    -400000 200000 -400000 200000
    66666.67 -133333.33 66666.67 -133333.33

    I took a=-400000, b=200000, c=66666.67, d= -133333.33
  8. Sep 19, 2013 #7


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    Take the simpler example of a = b = c = d = 0.

    There is obviously something wrong with the question here.
  9. Oct 5, 2013 #8
    Switching around the rows you have A' = ##
    a & b & a & b\\
    c & d & c & d\\
    0 & 0 & 1 & 0\\
    0 & 0 & 0 & 1\\
    \end{pmatrix}## and B = ##\begin{pmatrix}
    0 & 0 & 0 & 0\\
    0 &0 & 0 & 0\\
    0 & 0 & a & b\\
    0 & 0 & c & d\\

    Clearly A' has two eigenvalues of 1 -- they are sitting right there on the diagonal; Since A' was obtained by switching each row and even number of times, the eigenvalues of A' are those of A. Clearly also B has two eigenvalues of 0. What are the other eigenvalues of B? If b = c = 0 then they are a and d. If b and c are 0 a and d will also be the eigenvalues of A. You will want to show this, but the computation should be easy with all those zeros in it.

    However, b and c don't have to be 0. My guess would be that the 2 eigenvalues of ##\begin{pmatrix}
    a & b\\
    c & d\\
    \end{pmatrix}## will also be the other two eigenvalues of A.

    Can you show that?
    Last edited: Oct 5, 2013
  10. Dec 20, 2013 #9
    In my case a,b,c,d are not zeros alephzero. Thanks for the reply
  11. Dec 20, 2013 #10
    Thanks brmath. That helps
  12. Jan 13, 2014 #11
    Can we obtain relation between eigenvectors of A and B matrices
  13. Jan 13, 2014 #12
    Will try to get back to you later today.
  14. Jan 13, 2014 #13


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    Take brmath's re-arranged matrix A' and consider vectors of the form (x,y,0,0).
  15. Jan 13, 2014 #14
    Thanks for the reply.I did not understand what u meant by consider vectors of the form (x,y,0,0). I already tried using A' matrix to solve it but could not go any further.
  16. Jan 13, 2014 #15
    For the example that I took A has eigenvectors
    [2.206e-6 6.008e-6 0.6912 -0.3835;
    -4.749e-7 9.304e-6 -0.1487 -0.5940;
    -0.9776 -0.5424 -0.6912 0.3835;
    0.2104 -0.84007 0.1487 0.5940]
    and B has
    [0 0 1 0;
    0 0 0 1;
    -0.9776 -0.5424 0 0;
    0.2104 -0.84007 0 0].

    Eigenvectors of [a b; c d] is
    [-0.9776 -0.5424;
    0.21043 -0.84007]
  17. Jan 13, 2014 #16
    The eigenvectors present a different kind of problem, and there are a number of different possibilities.

    I suggest you start by finding the eigenvectors of B which match the 0 eigenvalues: i.e. Bx = 0. You will get either one or two different x's. I suspect just one.

    With the A there are numerous possibilities, which depend on the a, b, c, and d. For example if a = c = 1 and b = d = 0, A will probably have four independent eigenvectors all corresponding to the single eigenvalue 1. If you have a = b = d = 1 and c = 0, A will probably have 3 eigenvectors corresponding to the eigenvalue 1. You will have to work this out.

    In either of these cases, B will also have eigenvectors corresponding to 1 - -either two independent ones, or just one.

    Whether any of these eigenvectors match up between A and B is something you will have to compute. That is, find the eigenvectors of A which correspond to 1 under the two a,b,c,d scenarios I suggested, and find the eigenvectors of B for those same 1's.

    Offhand I see no particular reason to believe they are the same or different -- you'll have to see.

    Now the x's that match with the zero eigenvalues of B might or might not be eigenvalues of A. It could be that they are for some values of a,b,c,d and likely not for others. But you should check by multiplying the x's by A.

    Once you've gotten through all that, you may have a clue as to whether anything matches up for other values of a,b,c,d.
  18. Jan 13, 2014 #17


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    Calculate A'(x,y,0,0)t and you should notice that it looks a lot like a 2x2 matrix operating on a two dimensional vector.
  19. Jan 14, 2014 #18
    It does, but I think the a,b,c,d create a lot of complications.
  20. Jan 20, 2014 #19
    A and B are block matrices. So are xI-A and xI-B. Does that give you a way to find their characteristic polynomials? If I have to guess:
    charpoly(B,x) = x^2((x-a)(x-d) -bc) and charpoly(A,x)=(x-1)(x-1)((x-a)(x-d) -bc)
    so both charpoly(A) and charpoly(B) have common (quadratic) factors ((x-a)(x-d) -bc) .
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