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How to find eigenvalues/eigenvectors

  1. Mar 4, 2008 #1
    How do i find the eigenvalues and eigenvectors for the linear operator T defined as
    T(w,z) = (z,w)??
  2. jcsd
  3. Mar 4, 2008 #2
    I'd start by writing T as a matrix.
  4. Mar 5, 2008 #3
    how about for something like: T(x_1,x_2,...,x_n) = (x_1+x_2+...+x_n, x_1+x_2+...+x_n, ..., x_1+x_2+...+x_n). The matrix with respect to standard basis would have 1's everywhere? any clues to finding the eigenvalues/vectors?
  5. Mar 5, 2008 #4
    Try Matlab command >>[V,D] = eig(ones(n))
  6. Mar 5, 2008 #5


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    ??? T operates on a pair of numbers and gives a pair of numbers as the result. Written as a matrix, it would be 2 by 2 matrix- certainly not as complicated as you have! You are not still referring to the first problem are you?

    By definition, T(w,z)= (z, w) so T(1, 0)= (0, 1) and T(0, 1)= (1, 0).
    [tex]\left(\begin{array}{cc} a & b \\ c & d \end{array}\right)\left(\begin{array}{c} 1 \\ 0\end{array}\right)= \left(\begin{array}{c} 0 & 1\end{array}\right)[/tex]
    What are a and c? Do the same with (0, 1) being taken to (1, 0) to determine b and d.
    A good way of determining the matrix representing a linear operator in a given basis is to apply it to each of the basis vectors in turn. The result will be a column of the matrix.

    Of course, you don't have to write it as a matrix to find eigenvalues- in fact, I think too many students get the idea that Linear Algebra is only about matrices. Saying that [itex]\lambda[/itex] is an eigenvalue for linear transformation T means that there exist some (x, y), not both 0, such that [itex]T(x,y)= \lambda(x, y)= (\lambda x, \lambda y)[/itex]. Since T(x,y)= (y, x), that says that [itex](y, x)= (\lambda x, \lambda y)[/itex] so you have two equations: [itex]y= \lambda x[/itex] and [itex]x= \lambda y[/itex]. Obviously, x= y= 0 would satisfy those equations for any [itex]\lambda[/itex]. For what values of [itex]\lambda[/itex] would that have non-zero solutions? If you replace the "x" in the first equation by [itex]\lambda y[/itex] from the second equation, you have [itex]y= \lambda(\lambda y)= \lambda^2 y[/itex]. If y is not 0, you can divide both sides by y to get [itex]\lambda^2= 1[/itex].
  7. Mar 5, 2008 #6


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    One of the things you should have learned long ago is that you approach problems like this by looking at simple cases: if n= 2, this says T(x,y)= (x+ y, x+ y). In particular, T(1, 0)= (1, 1) and T(0,1)= (1, 1). Yes, the columns of the matrix representing this linear operator in the standard basis are all 1s. The matrix representing this linear operator in the standard basis consists of all 1s.

    Okay, what are the eigenvalues of
    [tex]\left(\begin{array}{cc}1 & 1 \\ 1 & 1\end{array}\right)[/tex]?
    (Hint: if a matrix has two rows (or two columns) the same it has determinant 0. And if it has deteminant 0, it has 0 as an eigenvalue.)

    The eigenvalues must satisfy
    [tex]\left|\begin{array}{cc} 1-\lambda & 1 \\ 1 & 1- \lambda\end{array}\right|= 0[/tex]

    What equation does that give you? What are the eigenvalues?
  8. Mar 5, 2008 #7


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    it seems obvious that (1,1) goes to (1,1), and (1,-1) goes to???? so the eigenvalues are...
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