# Frame operator on finite dimensional Hilbert space.

A sequence of distinct vectors $\{f_1,f_2,...\}$ belonging to a separable Hilbert space $H$ is said to be a Frame if there exist positive contants $A$, $B$ such that, for $A<B$: $$A\|f\|^2\leq\sum_{n=1}^\infty |(f,f_n)|^2\leq B \|f\|^2.$$ An operator $T$ on $H$ is said "frame operator" of frame $\{f_1,f_2,...\}$, if $$Tf=\sum_{n=1}^\infty (f,f_n) f_n$$ It is known that this operator, on infite dimensional Hilbert space, has various properties. For example, it is bounded, positive, self-adjoint. Are these properties preserved also for frames on finite dimensional Hilbert space?

## 1 Answer

Yes, just note that $(Tf,f) = \sum_{n=1}^N \vert (f,f_n) \vert^2$, where $N$ is the (possibly infinite) number of frame vectors, and that this quantity lies between $A \Vert f \Vert^2$ and $B \Vert f \Vert^2$. This shows that $T$ is bounded with $\Vert T \Vert \le B$, and selfadjoint (because $(Tf,f) \in \mathbb{R}$ for all $f$), and positive (because $(Tf,f) \ge 0$ for all $f$).

Moreover, the frame operator is also invertible because of the lower bound, and $\Vert T^{-1} \Vert \le \frac{1}{A}$. The wikipedia page concerning frames is not bad. You should have a look if you want more information than I provided above.