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Q1: Does A and its transpose have the same eigenspace?

  1. Feb 15, 2008 #1
    So I've shown that A and A^T have the same char. polynomials => same eigenvalues, using the fact that detA = detA^T. I still can't see any way I could possibly show or disprove that the eigenspace is the same.
     
  2. jcsd
  3. Feb 15, 2008 #2

    gel

    User Avatar

    How about
    [tex]
    \left(\begin{array}{ll}
    1 & 1\\
    0 & 0
    \end{array}\right)\ \
    \left(\begin{array}{ll}
    1 & 0\\
    1 & 0
    \end{array}\right)
    [/tex]
    Do they have the same eigenspaces?
     
  4. Feb 15, 2008 #3
    A and A^T will not have the same eigenspaces, i.e. eigenvectors, in general.

    Remember that there are in fact two "eigenvectors" for every eigenvalue [tex]\lambda[/tex]. The right eigenvector satisfying [tex]A\mathbf{x} = \lambda \mathbf{x}[/tex] and a left eigenvector (eigenrow?) satisfying [tex]\mathbf{x}A = \lambda \mathbf{x}[/tex]. In general these are not equal.

    Also, I believe that the set of left eigenvectors is the inverse matrix of the set of right eigenvectors, but I am not about sure of this. If this is indeed the case then the set of left eigenvectors will "coincide" with the set of right eigenvectors only when the set of right eigenvectors is orthonormal, i.e. when A is symmetric A=A^T.

    EDIT: In fact, the conjecture above is not true unless you select specially scaled sets of eigenvectors and eigenrows. Is there a way of selecting eigenvectors of canonical lengths?
     
    Last edited: Feb 15, 2008
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