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every orthonormal basis is a parseval frame. but what about the converse in the finite dimensional case?
Let's say $H$ is a n-dimensional Hilbert space and $a_1,..,a_n$ a parseval frame. then, of course, $a_1,..,a_n$ is a basis, since every frame spans the whole space (or maybe the closure in the infinite dimensional case).
But is it also an orthonormal basis then? I mean it satisfies Parsevals identity by definition. Does anybody know how to prove or contradict that?

Greatings
Roman

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  • $\begingroup$ Not every Parseval frame is an o.n. basis, see en.wikipedia.org/wiki/Frame_of_a_vector_space#Tight_frames $\endgroup$
    – Yemon Choi
    Jan 7, 2015 at 0:10
  • $\begingroup$ Oh, I see that you were only asking about frames where the number of elements is the same as the dimension of the space. In that case @Flounderer's answer solves your question $\endgroup$
    – Yemon Choi
    Jan 7, 2015 at 0:19

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Yes, because a Parseval frame has the property that for any $i$, $$a_i = \sum_{j=1}^n \langle a_i, a_j \rangle a_j$$ and since the $a_i$ form a basis, the coefficients of $a_i$ on both sides must be the same, which implies that all the $a_i$ are orthogonal to one another, and also that $||a_i||^2 = 1$.

(If you are starting from the definition $||x||^2 = \sum |\langle x, e_i \rangle |^2$ for all $x$, it is a standard fact that this is equivalent to $x = \sum \langle x, e_i\rangle e_i$ for all $x$.)

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  • $\begingroup$ ah, the perfect reconstruction of parseval frames. i've seen that before. thank you! nice and easy explanation. =) $\endgroup$
    – Roman
    Jan 7, 2015 at 11:50

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