# finite spanning set<->frame

• March 3rd 2010, 06:55 PM
alighieri
finite spanning set<->frame
Show that every finite spanning set for $\mathbb{R}^d$ is a frame for $\mathbb{R}^d$
(frame: $\exists A,B >0 \in such that A ||x||^2<=\sum||^2<=B||x||^2$ )

and that every finite frame ${f_1,\ldots,f_n}\subset \mathbb{R}^d, n>d$ is a spanning set for $\mathbb{R}^d$
• March 4th 2010, 05:17 AM
verdi
finite frame->spanning

Assume ${f_1,\ldots,f_n}$ is a frame, but not a spanning set $\rightarrow$ nontrivial null space for the matrix with rows $f_k\rightarrow\exists x$ such that $||x||>0$ but $\forall k, ||=0\rightarrow$ Contradiction of framing condition for any postive A
• March 4th 2010, 05:26 AM
alighieri
Thanks. Also, half of the other direction is easy:

$\sum ||^2\leq\sum ||x||^2||f_k||^2$ by Cauchy-Schwartz $=||x||^2\sum||f_k||^2$ and spanning $\rightarrow \exists k$such that $||f_k||>0\rightarrow B=\sum||f_k||^2$