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Matrix-valued analytic function?

  1. Jun 25, 2013 #1
    Consider the matrix-valued function [itex]T(\beta): \mathbb C \to \mathscr L(\mathbb C^2)[/itex], the bounded linear operators on [itex]\mathbb C^2[/itex], given by

    [tex]
    T(\beta) = \begin{bmatrix} 1 & \beta \\ \beta & -1 \end{bmatrix}.
    [/tex]

    According to Reed-Simon, Volume 4, this is a "matrix-valued analytic function" with singularities at [itex]\beta = \pm i[/itex]. I'm confused as to how...
    1. ...we are supposed to show that [itex]T(\beta)[/itex] is analytic. The claim made in the book is that it is easier (in general) to show that a vector-valued analytic function is weakly analytic than strongly analytic, but I don't see how that is the case here.
    2. ...we are supposed to see that this function has singularities at [itex]\pm i[/itex].
    Can anyone help with either of the above? Thanks!
     
  2. jcsd
  3. Jun 25, 2013 #2
    Update

    It seems [itex]T(\beta)[/itex] is actually an entire matrix-valued analytic function of [itex]\beta[/itex]; it is the eigenvalues [itex]\lambda_\pm (\beta) = \pm \sqrt{\beta^2 + 1}[/itex] that have singularities at [itex]\pm i[/itex]. My question still stands, though...why are [itex]\pm i[/itex] singularities of this function? What's wrong with taking the square root of zero, which is what one winds up doing at those values?
     
  4. Jun 25, 2013 #3

    pasmith

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    Homework Helper

    Nothing. But there is a problem with evaluating the derivative of the square root function at zero, and that's what causes [itex]\sqrt{\beta^2 + 1}[/itex] to fail to be analytic at [itex]\beta = \pm i[/itex].
     
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