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Homework Help: Quick question on eigenvalues

  1. Mar 16, 2008 #1
    Let's say that I have to construct a 2 X 2 matrix from a second-order differential equation, turning it into a system of first order linear equations, and find its eigenvalues. I'll have two variables that correspond to the two columns in the matrix.

    If I swap columns, I end up with two different characteristic polynomials, and thus different sets of eigenvalues (this is the problem). How do I know which variable should correspond to the first column, and which variable should correspond to the second column?
  2. jcsd
  3. Mar 17, 2008 #2
    If you switch the columns, you switch the order of the variables.
  4. Mar 17, 2008 #3


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    You spent a course learning linear arithmetic, so use it. :wink:

    Swapping columns is a matrix operation: if you started with the equation

    [tex]A x = y[/tex]

    and you want to replace A with AC, where C is the column swap elementary matrix... i.e.

    C = \left(
    0 & 1 \\
    1 & 0

    Can you think of any way to modify the equation Ax=y (without changing its validity!) to make a C appear somewhere in it? (Preferably just to the right of the A)
  5. Mar 17, 2008 #4
    Thanks for the replies :)

    My math professor helped me out this morning, and I found out that if I have a second order differential equation corresponding to an equation y(t), and I let dy/dt = v, then the "rule of thumb" is to let the coefficient matrix A have its first column correspond to the variable y and let the second column correspond to the derivative of y(t), v (this way I know which coefficients in A to subtract [tex]\lambda[/tex] from). But now I want to know what you're talking about! So I'll take a shot at your question :)

    I believe your question is, how can I place the column swap matrix C into the equation Ax = y, without the equation losing its validity. To do this, I believe I would multiply both sides of Ax = y by C, such that C appears between A and x, and to the left of y. So I would have ACx= Cy. I worked this out on paper and I see that this does indeed swap the columns in A and the rows in y, which wouldn't change my answer, so it works out. Please let me know if I'm mistaken.
  6. Mar 17, 2008 #5


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    That doesn't quite work -- remember that the operations you have available are "left multiply" and "right multiply"... so you can't just insert it anywhere you want.

    The method I was hinting at is this idea: If I want to right-multiply A by C, I can undo that by right-multiplying by C-inverse. In this case, the inverse of C is itself, so I have the following derivation:

    Ax = y
    A (CC) x = y
    (AC) (Cx) = y

    The procedure you used shouldn't work in general: if Ax=y, then usually ACx=Cy will be false. In fact, that implies:

    ACx = Cy
    ACx = C(Ax)
    ACx = CAx
    (AC - CA) x = 0

    so it will only work when the solution for x happens to be a nullvector of (AC - CA).
  7. Mar 17, 2008 #6
    I see, so this works out because C*C = I, where I would be the 2 X 2 identity matrix, and AI = A.
  8. Mar 17, 2008 #7


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    Right. And if you did a more complicated column operation to A, you can hopefully work out what happens to x to neutralize it.
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