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Continuous square root function on the space of nxn matrices

  1. Dec 22, 2009 #1
    1. The problem statement, all variables and given/known data
    Hi again all,
    I've just managed to prove the existence (non-constructively) of a 'square root function' f on some open epsilon-ball about the identity matrix 'I' such that [itex][f(A)]^2=A\qquad \forall\, A \,\text{ s.t.}\, \|I-A\|<\epsilon[/itex] within Mn, the space of n*n matrices (note that's f(A)^2, not f^2(A), so for example the identity function wouldn't work) - I used the inverse function theorem on A^2 to deduce its existence. However, I was wondering whether there exists a function f such that f2(A)=A [itex]\forall A \in M_n[/itex]? Or does there only exist such a function in a finite ball? What about for something like a cube root or a quintuple root function?

    I was thinking perhaps some sort of compactness argument might work, but I couldn't reason anything in particular (and that's only if it isn't possible for the whole of [itex]M_n[/itex], otherwise compactness wouldn't work I don't suppose)

    Many thanks for any help :)
     
    Last edited: Dec 22, 2009
  2. jcsd
  3. Dec 23, 2009 #2
    Any thoughts, anyone? Any help at all would be appreciated!
     
  4. Dec 23, 2009 #3
    This is actually a fairly substantial question. Not every matrix has a square root; for instance, [tex]\begin{pmatrix}0&1\\0&0\end{pmatrix}[/tex] does not. But positive semidefinite matrices do; there are several possible constructions. The most "fundamental" one from a functional analysis point of view uses something called the functional calculus, which makes it possible to "apply" many real functions, not just the square root function, to linear operators.

    Look at the Wikipedia article "Square root of a matrix" to start. For a more pedagogical treatment, there is lots of good material about spectral theory in finite-dimensional spaces (which is what this is related to) in Paul Halmos's book Finite-dimensional vector spaces.
     
  5. Dec 23, 2009 #4
    Thanks ever so much, how stupid of me not to spot that! Is it true of every order then, that there exists some matrix which isn't the n'th power of any matrix? (I couldn't come up with an example this late at night for the cube root, but I may be being slow ;-))
     
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