ChapterZero

I spent way too much time trying to remember why every linear function on a finite-dimensional vector space is bounded. (because if and , then , so ).

I need this result to show that every finite spanning set in a finite dimensional Hilbert space is a frame. That is, if spans then there are such that

for each .

The function defined by is linear, so by the above result is bounded with some constant .

Now I just need show that the inverse is continuous, so there is a …

This problem came up when I was talking to a fellow graduate just now about what I’m doing in Papadakis’ group, and I tried introducing the concept of a frame. The first question he asked was if each spanning set is not necessarily a frame. It seems that this is true in the finite dimensional case, but I’m sure it’s not in general Hilbert spaces, because then what would be the point of defining a frame? I’ll see if I can find a spanning set that isn’t a frame in .

Update:
The lower bound exists because the inverse function is again a linear function on a finite dimensional vector space, so it is bounded by some constant . We can take .

It is easy to find a spanning set in that is not a frame, because it is not a Bessel sequence (a sequence which defines a bounded operator ): simply let .

Possibly relevant posts:

• Inner products (6/7/2006)
• Not a frame, but Bessel (5/26/2006)
• I’m stumped– you give it a try (4/1/2008)