In a finite dimensional vector space a positive operator can have infinitely many square roots "In a finite dimensional vector space a identity operator (which is positive) can have infinitely many square roots". I see only $2 ^ {C(n,2)}$ when dimension is $n (>1)$ by swapping two basis elements. Where am I wrong ?
 A: Let $X: \mathbb{R} \to \mathbb{R}^{2 \times 2}$,
$$X(t) := \begin{bmatrix} 1 & t \\ 0 & -1 \end{bmatrix}.$$
Then
$$(X(t))^2 := \begin{bmatrix} 1 & t \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & t \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$$
for all $t \in \mathbb{R}$. I leave it to you to generalize this example to higher dimension spaces.
Note that these are merely all triangular matrices $X$ such that $X^2 = I$. There are many other, nontriangular ones, with the same property.
A: To elaborate on your example, $Y=\pmatrix{0&1\\ 1&0}$ is a square root of $I$. Therefore, $X=PYP^{-1}$ is also a square root of $I$ for every invertible $P$. One can obtain infinitely many such $X$s by using different $P$s. And basically, all square roots of $I$ are obtained in this manner, i.e. they are similar to a diagonal matrix whose diagonal entries (or eigenvalues) are $\pm1$.
A: First of all, a positive operator and an identity operator are much different.
Second of all, the square root of an identity is an identity, and so forth.
