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 $n (&gt;1)$ by swapping two basis elements. Where am I wrong ?
Let $X: \mathbb{R} \to \mathbb{R}^{2 \times 2}$,

$$X(t) := \begin{bmatrix} 1 &amp; t \\ 0 &amp; -1 \end{bmatrix}.$$

Then

$$(X(t))^2 := \begin{bmatrix} 1 &amp; t \\ 0 &amp; -1 \end{bmatrix} \begin{bmatrix} 1 &amp; t \\ 0 &amp; -1 \end{bmatrix} = \begin{bmatrix} 1 &amp; 0 \\ 0 &amp; 1 \end{bmatrix} = I$$

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.
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.