# 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 $\beta = \pm i$.