Notation $2^T$ for the operator T of the Hilbert space I came across an unknown to me notation $2^T, 2^{-T}$ for a positive operator $T$ on a Hilbert space (more precisely, in $\mathbb{C}^n$).
What does this mean? I can't find it anywhere. Is it the image and the kernel?
 A: For a positive (more generally, normal) operator $T$ on a Hilbert space $H$, there is a unique unital *-homomorphism (a linear map that preserves products and involutions) from $C(\sigma(T))$ to $B(H)$ that maps $\operatorname{id}$ to $T$, where $\operatorname{id}$ is the map $C(\sigma(T))\to \mathbb{C}, z\mapsto z$.
For a continuous function $f$ on the spectrum $\sigma(T)$ of $T$, $f(T)$ means the image of $f$ under the unique *-homomorphism.
For example, let $p(x)=x^2+2x+1$,  you can check $p(T)=T^2+2T+e$ where $e$ is the  identity of $B(H)$.
For $f(x)=2^x$, it can be approximated by polynomials $p_N(x)=\sum_{n=0}^N\frac{(\ln 2)^n}{n!}x^n$ with respect to the uniform norm. $f(T)$ can be proved to be the limit of $p_N(T)$.
A: Since $T$ is a positive operator, it is a normal operator and we can consider the continuous functional calculus
$$C(\sigma(T)) \to  B(\mathbb{C}^n) \cong M_n(\mathbb{C})$$
Then $2^T$ is the operator obtained as the image of $\sigma(T)\ni x \mapsto 2^x$ under the above map.
In some sense, it is unnatural to consider the operator $2^T$ since $\sigma(T)$ is a finite space, but I think it suffices for your purposes.
