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Find a 2x2 Matrix with given EigenValues

  1. Jul 23, 2015 #1
    1. The problem statement, all variables and given/known data
    Find a 2X2 matrix that has all non-zero entries where 3 is an eigenvalue

    2. Relevant equations


    3. The attempt at a solution
    well since the 2x2 matrix cannot be triangular, it makes things harder for me.
    I have no idea where to start. I am not given any eigenvectors either.
    It seems like a simple problem, but I've been stuck on it for a while.

    If I did not have the zero entries restriction i would have selected 2x2 matrix
    A= {{3,0}{0,3}}τ

    Can Someone point me in the right direction?
     
  2. jcsd
  3. Jul 23, 2015 #2

    ShayanJ

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    Gold Member

    The eigenvalue equation is for the 2X2 matrix, if written as a system of homogeneous equations, will have a solution if the determinant of the matrix of coefficients is zero. So we have the equation ## \lambda^2-(a+d)\lambda+ad-bc=0## where ## \lambda ## is the given eigenvalue and a,b,c and d are the unknown matrix entries. So we have one equation for four unknowns which means this is an under-determined system. So there are an infinite number of 2X2 matrices with a given eigenvalue. Just choose three of the entries arbitrarily and determine the fourth via the above equation.
     
  4. Jul 23, 2015 #3
    thanks for your help Shyan Using that system I was able to solve my problem :)
     
  5. Jul 24, 2015 #4

    HallsofIvy

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    First, while you said that "3 is an eigenvalue", you did NOT say that it was the only eigenvalue, a double eigenvalue. Even assuming that there are an infinite number of such 2 by 2 matrices. One way to find one is to start with the diagonal matrix [tex]M= \begin{bmatrix}3 & 0 \\ 0 & 3\end{bmatrix}[/tex]. Now, for any invertible matrix, A, [itex]AMA^{-1}[/itex] will have the same eigenvalues so you only need to choose A so that has no 0 entries.

    Completely different solutions can be found by starting, not with a diagonal matrix, but with the "Jordan Normal Form", [tex]M= \begin{bmatrix}3 & 1 \\ 0 & 3\end{bmatrix}[/tex]
     
  6. Jul 24, 2015 #5

    pasmith

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    Homework Helper

    If [itex]M[/itex] is a multiple of the identity, then it commutes with [itex]A[/itex] so that [itex]AMA^{-1} = MAA^{-1} = M[/itex], which has zero entries.

    This is necessary if 3 is to be a repeated eigenvalue, but the question doesn't require that.
     
  7. Jul 25, 2015 #6

    HallsofIvy

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    Good points- thank you.
     
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