The Kadison-Singer and Paulsen Problems in Finite Frame Theory

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22 Peter G. Casazza2N∑i=1|〈x, ϕ i 〉| 2 =(1 − ɛ)22N∑|〈x, e i 〉| 2 +i=1(1 + ɛ)22N∑|〈x, e i 〉| 2 = (1 + ɛ 2 )‖x‖ 2 .i=1So {ϕ i } 2Ni=1 is a (1 + ɛ2 ) tight frame and hence an ɛ-nearly Parseval frame.The closest equal norm frame to {ϕ i } 2N eii=1 is { √2} N i=1 ∪ { √ ei2} N i=1 . Also,N∑i=1‖ e i√ −ϕ i ‖ 2 + 22N∑i=N+1‖ e i−N√ −ϕ i ‖ 2 = 2N∑i=1‖ ɛ √2e i ‖ 2 +‖N∑i=1‖ ɛ √2e i ‖ 2 = ɛ 2 N.The main difficulty with solving the Paulsen Problem is that finding aclose equal-norm frame to a given frame involves finding a close frame whichsatisfies a geometric condition while finding a close Parseval frame to a givenframe involves satisfying a (algebraic) spectral condition. At this time, welack techniques for combining these two conditions. However, each of themindividually has a known solution. That is, we do know the closest equalnorm frame to a given frame [15] and we do know the closest Parseval frameto a given frame [5, 10, 15, 22, 35].Lemma 3. If {ϕ i } M i=1 is an ɛ-nearly equal norm frame in HN , then the closestequal norm frame to {ϕ i } M i=1 iswhereψ i = a ϕ i, for i = 1, 2, . . . , M,‖ϕ i ‖∑ Mi=1a =‖ϕ i‖.MIt is well known that for a frame {ϕ i } M i=1 for HN with frame operator S,the closest Parseval frame to {ϕ i } M i=1 is {S−1/2 ϕ i } M i=1 [5, 10, 15, 22, 35]. Wewill give the version from [10] here.Proposition 5. Let {ϕ i } M i=1 be a frame for a N-dimensional Hilbert spaceH N , with frame operator S = T ∗ T . Then {S −1/2 ϕ i } M i=1 is the closest Parsevalframe to {ϕ i } M i=1 . Moreover, if {ϕ i} M i=1 is an ɛ-nearly Parseval frame thenM∑‖S −1/2 ϕ i − ϕ i ‖ 2 ≤ N(2 − ɛ − 2 √ 1 − ɛ) ≤ Nɛ 2 /4.i=1Proof. We first check that {S −1/2 ϕ i } M i=1 is the closest Parseval frame to{ϕ i } M i=1 .The squared l 2 -distance between {ϕ i } M i=1 and any Parseval frame {ψ j} n j=1can be expressed in terms of their analysis operators T and T 1 as‖F − G‖ 2 = T r[(T − T 1 )(T − T 1 ) ∗ ]= T r[T T ∗ ] + T r[T 1 T ∗ 1 ] − 2RT r[T T ∗ 1 ]
1 The Kadison-Singer and Paulsen Problems in Finite Frame Theory 23Choosing a Parseval frame {ψ i } M i=1 is equivalent to choosing the isometry T 1.To minimize the distance over all choices of T 1 , consider the polar decompositionT = UP , where P is positive and U is an isometry. In fact, S = T ∗ Timplies P = S 1/2 , which means its eigenvalues are bounded away from zero.Since P is positive and bounded away from zero, the term T r[T T ∗ 1 ] =T r[UP T ∗ 1 ] = T r[T ∗ 1 UP ] is an inner product between T 1 and U. Its magnitudeis bounded by the Cauchy Schwarz inequality, and thus it has a maximal realpart if T 1 = U which implies T ∗ 1 U = I. In this case, T = T 1 P = T 1 S 1/2 ,or equivalently T ∗ 1 = S −1/2 T ∗ , and we conclude ψ i = S −1/2 ϕ i for all i =1, 2, . . . M.After choosing T 1 = T S −1/2 , the l 2 -distance is expressed in terms of theeigenvalues {λ j } N j=1 of S = T ∗ T by‖F − G‖ 2 = T r[S] + T r[I] − 2T r[S 1/2 ]N∑N∑ √= λ j + N − 2 λj .j=1If 1 − ɛ ≤ λ j ≤ 1 + ɛ for all j = 1, 2, . . . N, calculus shows that the maximumof λ j − 2 √ λ j is achieved when λ j = 1 − ɛ.Consequently,j=1‖F − G‖ 2 ≤ 2N − Nɛ − 2N √ 1 − ɛ.Estimating √ 1 − ɛ by the first three terms in its decreasing power series givesthe inequality ‖F − G‖ 2 ≤ Nɛ 2 /4.It can be shown that the estimate above is exact and so we have separateverification that the closeness function is a function of N.There is a simple algorithm for turning any frame into an equal normframe with the same frame operator due to Holmes and Paulsen [34].Proposition 6. There is an algorithm for turning any frame into an equalnorm frame with the same frame operator.Proof. Let {ϕ i } M i=1 be a frame for HN with frame operator S and analysisoperator T . ThenM∑‖ϕ i ‖ 2 = T r S.i=1Let λ = T r SM . If ‖ϕ i‖ 2 = λ, for all m = 1, 2, . . . , M, then we are done.Otherwise, there exists 1 ≤ i ≠ j ≤ M with ‖ϕ i ‖ 2 > λ > ‖ϕ j ‖ 2 . For any θ,replace the vectors ϕ i , ϕ j by the vectorsψ i = (cosθ)ϕ i − (sinθ)ϕ j , ψ j = (sinθ)ϕ i + (cosθ)ϕ j , ψ k = ϕ k , for k ≠ i, j.Now, the analysis operator for {ψ i } M i=1 is T 1 = UT for a unitary operator Uon l N 2 given by the Givens’ rotation. Hence, T1 ∗ T 1 = T ∗ U ∗ UT = T ∗ T = S; so
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