Square root of positive operator In my linear algebra textbook (Roman, pg 252) it's stated that "Since $\tau$ is a positive operator, it has a positive square root $\sqrt{\tau}$, which is a polynomial in $\tau$. A similar statement holds for $\sigma$. Therefore, since $\tau$ and $\sigma$ commute, so do $\sqrt{\tau}$ and $\sqrt{\sigma}$." 
Why is $\sqrt{\tau}$ a polynomial in $\tau$? $\tau$ is a polynomial in $\sqrt{\tau}$, but I don't see why $\sqrt{\tau}=p(\tau)$.
 A: Suppose $\{\lambda_1,\ldots,\lambda_n\}$ is the spectrum of $\tau$. Let $p$ be any polynomial (such as a Lagrange interpolation polynomial) such that $p(\lambda_i)=\sqrt{\lambda_i}$ for each distinct $\lambda_i$. Then $p(\tau)=\sqrt{\tau}$.
A: I assume that you're considering the finite dimensional case. In that case you can use Jordan Normal form to see this. Also we use that powers of the Jordan Normal form will be a matrix with blocks being powers of the jordan blocks. So we really only have to consider Jordan blocks.
Now a Jordan block is on the form $J = \lambda I + S_1$ (where $S_1$ is one-step shift matrix). Also we have that $\lambda>0$. 
Now we shall consider the Taylor expansion of the square root around $\lambda$:
$\sqrt{\lambda + x} = \sum_0^n c_j x^j + o(x^n)$
This means that $(\sum_0^n c_jx^j)^2 + o(x^{2n}) = \lambda+x$, but since the sum is a polynomial and the RHS is a polynomial the term $o(x^{2n})$ must be a polynomial $p_{2n+1}$ in $x$ having the lowest term $x^{2n+1}$. Since this is true in a neighborhood of $x$ (by Laurin theorem) it's also true universally (since it's a polynomial identity).
Now back to the Jordan block. Now we therefore have:
$$(\sum_0^n c_j S_1^j)^2 = \lambda I + S_1 - p_{2n+1}(S_1)$$
But since the one-step shift when repeated will eventually annihilate every vector that is $S_1^{2n+1}=0$ for some $n$. Using that $n$ in the above expression we have an expression for the square root of the Jordan block:
$$\sqrt{J} = \sqrt{\lambda I + S_1} = \sum_0^n{c_j S_1^j}$$
Next step is to see how we can express the RHS as a polynomial in $J$. We see this simply by examining the binomial theorem stating that $J^k$ is a polynomial in $S_1$ of degree $k$. We then can select $a_k J^k$ such that the $S_1^k$ term matches that of $\sqrt{J}$ and then continue the process and finally reach a polynomial in $J$ that matches the polynomial $\sum_0^n{c_j S_1^j}$.
Note that it is crucial that $\tau$ is strictly positive. If it have zero eigenvalues this approach does not work. 
