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Linear operators, eigenvalues, diagonal matrices

  1. Mar 9, 2010 #1
    So I have a couple of questions in regards to linear operators and their eigenvalues and how it relates to their matrices with respect to some basis.

    For example, I want to show that given a linear operator T such that [tex]T(x_1,x_2,x_3) = (3x_3, 2x_2, x_1)[/tex] then T can be represented by a diagonal matrix with respect to some basis of [tex]V = \mathbb{R}^3[/tex].

    So one approach is to use a theorem that says: If T has dim(V) distinct eigenvalues, then T has a diagonal matrix with respect to some basis V.

    So we simply look for the solutions of: [tex]\lambda x_1 = 3 x_3[/tex], [tex]\lambda x_2 = 2 x_2[/tex], [tex]\lambda x_3 = x_1[/tex]

    So we find that [tex]\lambda = 2, \pm \sqrt 3[/tex] with some corresponding eigenvectors.So we use the theorem and finish off the question.

    So finally, my problem with all this. The matrix of T with respect to the standard basis {(0,0,1), (0,1,0),(1,0,0)} (purposely out of order) would be:

    [tex]\begin{bmatrix} 3 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{bmatrix}[/tex]

    which is diagonal right? But I have a theorem that says that the eigenvalues of T with respect to a uppertriangular matrix "consist precisely of the entries on the diagonal of that upper-triangular matrix".

    But I didn't find 3 and 1 as eigenvalues. Is it because I switched the standard basis elements out of order? Why would that matter? And doesn't this last theorem imply that every linear operator precisely has dim V eigenvalues :S?

    It was a long one but any comments are appreciated!
  2. jcsd
  3. Mar 10, 2010 #2


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    This is wrong. If we call the basis vectors [itex]v_1,v_2,v_3[/itex] (in the order you've written them), we have for example

    [tex]T_{31}=(Tv_1)_3=(3v_3)_3=3(v_3)_3=3\cdot 1=3[/tex]

    The third component of [itex]v_3[/itex] in the basis you have defined is 1, not 0, because [itex]v_3=0\cdot v_1+0\cdot v_2+1\cdot v_3[/itex].
  4. Mar 10, 2010 #3


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    No, if you use that order in the "domain", you must also use it in the "range".
    T(0, 0, 1)= (3, 0, 0)= 0(0, 0, 1)+ 0(0, 1, 0)+ 3(1, 0, 0) so the first column will be
    [tex]\begin{bmatrix}0 \\ 0 \\ 3\end{bmatrix}[/tex]
    T(1, 0, 0)= (0, 0, 1)= 1(0, 0, 1)+ 0(0, 1, 0)+ 0(1, 0, 0) so the third column will be
    [tex]\begin{bmatrix} 1 \\ 0 \\ 0\end{bmatrix}[/tex]

    In this basis, the matrix corresponding to T is
    [tex]\begin{bmatrix}0 & 0 & 1 \\0 & 2 & 0\\ 3 & 0 & 0\end{bmatrix}[/tex]

  5. Mar 10, 2010 #4
    Ah what a silly mistake I forgot to write each vector as a linear combination of the basis elements.

    But then I'm still confused about the whole eigenvalue issue. I know I must be misinterpreting the theorem. The elements along the diagonal are "precisely" the eigenvalues of T. Doesn't this mean that all linear operators have exactly dimV eigenvalues? (Obviously not but I don't see why..)

    Thanks a lot!
    Last edited: Mar 10, 2010
  6. Mar 10, 2010 #5


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    The elements along the main diagonal of a diagonal matrix or a "Jordan Normal form" for a non-diagonalizable matrix are the eigenvalues of the matrix but in general, it is much harder to find the eigenvalues of a matrix than just looking at the main diagonal!

    It is, however, true that allowing for multiplicities and allowing complex numbers, that any n by n matrix has n eigenvalues- because its eigenvalue equation is an nth degee polynomial which can be factored into n linear factors over the complex numbers.

    An n by n matrix may not have n independent eigenvectors.
  7. Mar 10, 2010 #6
    Ah right. The entries along the diagonal of a upper triangular matrix need not be distinct.

    So then for example: [tex]T(x_1, x_2, x_3) = (2x_1, 2x_2, 3x_3)[/tex]

    We only have 2 eigenvalues here but with respect to the standard basis, we would still have a diagonal matrix just with repeated elements along the diagonal. Right?

    That clears things up a lot more. Thanks!
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