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Eigenvalues of transpose linear transformation

  1. Apr 3, 2017 #1
    1. The problem statement, all variables and given/known data
    If ##A## is an ##n \times n## matrix, show that the eigenvalues of ##T(A) = A^{t}## are ##\lambda = \pm 1##

    2. Relevant equations


    3. The attempt at a solution
    First I assume that a matrix ##M## is an eigenvector of ##T##. So ##T(M) = \lambda M## for some ##\lambda \in \mathbb{R}##. This means that ##M^t = \lambda M##. Then ##\det (M^t) = \det (M)##. So ##\det(M) = \lambda^n \det (M)##. But I can't seem to cancel the determinants since we don't know if ##M## is invertible or not. This is where I get stuck.
     
  2. jcsd
  3. Apr 3, 2017 #2

    fresh_42

    Staff: Mentor

    The statement is plain wrong: choose ##A=0## or any diagonal matrix ##diag(\lambda_1,\ldots ,\lambda_n)## with ##\lambda_i## of your choice. And these are only the simple counterexamples. Could it be, that ##A\cdot A^t=1## is the given condition?
     
  4. Apr 3, 2017 #3
    But if ##M^t = \lambda M##, and if ##\lambda = \pm 1##, wouldn't the symmetric and anti-symmetric matrices work?
     
  5. Apr 3, 2017 #4

    fresh_42

    Staff: Mentor

    What does this mean for a vector? A column vector is a multiple of a row vector? This would only mean ##n=1## and ##\lambda (M-1)=0##, i.e. ##\lambda = 0## or ##M=1##. Since the claim is obviously wrong, why bother an attempt to prove it? It cannot be proven. Perhaps I didn't understand your ##T##. To me it looked like ##T : A \longmapsto A^t##.
     
  6. Apr 3, 2017 #5
    I'll write out the problems exactly as it is written: Let ##T## be a linear operator on ##M_{n \times n} (\mathbb{R})## defined by ##T(A) = A^t##. Show that ##\pm 1## are the only eigenvalues of 1.

    The problem is from a standard textbook (Friedberg), so I don't think there's anything wrong with the problem
     
  7. Apr 3, 2017 #6

    fresh_42

    Staff: Mentor

    O.k., I thought it is about the eigenvalues of ##A##. My fault, sorry. Then you might look at ##T^2## for the solution.
     
  8. Apr 3, 2017 #7
    That ends up working well. Why doesn't the determinant approach work well? I want to know so that next time I come across a similar problem I won't spend time going own that path
     
  9. Apr 3, 2017 #8

    fresh_42

    Staff: Mentor

    I think I know it now. It's the same thing as my error arose from. ##M## is a vector, and its determinant isn't important. We should only look at the determinant of ##T##, which is clearly a regular mapping with ##\det(T)^2 = 1##. The equation you found isn't wrong, but as you stated correctly, we don't know anything about ##M##, only that ##M \neq 0## which isn't enough, if we pass to the determinant of ##M##. I mean this reduces all ##n^2## information about ##M## to a single number, and more, it distracts from the properties of ##T##.
     
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