Matrix-valued analytic function?

1. Jun 25, 2013

AxiomOfChoice

Consider the matrix-valued function $T(\beta): \mathbb C \to \mathscr L(\mathbb C^2)$, the bounded linear operators on $\mathbb C^2$, given by

$$T(\beta) = \begin{bmatrix} 1 & \beta \\ \beta & -1 \end{bmatrix}.$$

According to Reed-Simon, Volume 4, this is a "matrix-valued analytic function" with singularities at $\beta = \pm i$. I'm confused as to how...
1. ...we are supposed to show that $T(\beta)$ 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 $\pm i$.
Can anyone help with either of the above? Thanks!

2. Jun 25, 2013

AxiomOfChoice

Update

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

3. Jun 25, 2013

pasmith

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