Finite frames in finite-dimensional spaces

$\left\{ v_{1}, \ldots , v_{n}\right\}$ is a Bessel sequence in $\mathbf{F}^{d}$, its synthesis operator corresponds to multiplication by the matrix $R$ that has $v_{1}, \ldots , v_{n}$ as columns, and its analysis operator is $C = R^{*}$, the Hermitian of the matrix $R$.

The following are equivalent: (i) $\left\{ v_{1}, \ldots , v_{n}\right\}$ spans $\mathbf{F}^{d}$, (ii) $S = C^{*} C$ is positive definite, (iii) $\left\{ v_{1}, \ldots , v_{n}\right\}$ is a frame for $\mathbf{F}^{d}$. Further, in case these hold, the optimal frame bounds for $\left\{ v_{1}, \ldots , v_{n}\right\}$ are $\lambda_{1}, \lambda_{d}$, where $\lambda_{1}$ is the smallest eigenvalue of $S$ and $\lambda_{d}$ is the largest eigenvalue.

The following are equivalent: (i) $\left\{ v_{1}, \ldots , v_{n}\right\}$ is linearly independent, (ii) $G = C C^{*}$ is positive definite, (iii) $\left\{ v_{1}, \ldots , v_{n}\right\}$ is a Riesz sequence in $\mathbf{F}^{d}$.

Show that $\left\{ v_{1}, \ldots , v_{N}\right\}$ is a frame sequence, i.e., it is a frame for $\operatorname {span}\left\{ v_{1}, \ldots , v_{N}\right\}$.

Assume $\left\{ v_{1}, \ldots , v_{N}\right\}$ is linearly independent, and let $A, B$ be frame bounds for $\left\{ v_{1}, \ldots , v_{N}\right\}$ as a frame for its span. Show that

Exhibit a frame for $\mathbf{F}^{d}$ that contains infinitely many nonzero vectors.

Show that if $\left\{ x_{n}\right\}_{n \in I}$ is a frame for $\mathbf{F}^{d}$ and $\inf_{n \in I}\left\| x_{n}\right\| > 0$, then $I$ is finite.