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The Eigenvalues and eigenvectors of a 2x2 matrix

  1. Nov 15, 2009 #1
    1. The problem statement, all variables and given/known data

    Let B = (1 1 / -1 1)
    That is a 2x2 matrix with (1 1) on the first row and (-1 1) on the second.

    2. Relevant equations



    3. The attempt at a solution

    A)

    (1 1 / -1 1)(x / y) = L(x / y)

    L(x / y) - (1 1 / -1 1) (x / y) = (0 / 0)

    ({L - 1} -1 / 1 {L-1}) (x / y) = (0 / 0)

    Det (LI - B) = ({L - 1} -1 / 1 {L-1}) = 0

    ({L - 1} {L-1}) - (1)(-1)

    L^2 -2L +2 = 0

    L= 1 - i
    = 1+i

    So when L = 1-i

    ({1 -i - 1} -1 / 1 {1 -i -1})

    (-i -1 / 1 -i)

    -ix - y = 0
    x - iy = 0

    let x = t

    t - iy = 0
    y = t/i

    Im not sure if that even makes sense. Or how I would continue.

    B) Write the eigenvalues L of B in the form w = re^i(theta)

    If someone could just give me a little nudge in the right direction for this one because I dont even know where to start.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 15, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You are actually doing pretty well. You have the two eigenvalues right, and you've shown that an eigenvector of 1-i is given by x=t, y=t/i=(-it) for any nonzero value of t. That makes it t*(1 / -i). Now just do the same thing for 1+i. For the second part e^(i*theta)=cos(theta)+i*sin(theta). For a complex number L, the 'r' will be |L|. So L/|L|=cos(theta)+i*sin(theta). Just match up the real and imaginary parts and find theta.
     
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