MA342A (Harmonic Analysis 1) Tutorial sheet 3 [October 29, 2015] Name: Student ID: 1. For G = {g0 , g1 , . . . , gN −1 } an abelian group with N elements, let Ĝ = {χ0 , χ1 , . . . , χN −1 } be the dual group. Let A be the N × N matrix with (j, k) entry χj−1 (gk−1 ) (for 1 ≤ j, k ≤ N ). Show that AA∗ is N IN (where A∗ is the conjugate matrix transposed and IN is the N × N identity matrix). Solution: Each row of A are the N -tuple of values of one character. So row j has the values χj−1 (g0 ) χj−1 (g1 ) · · · χj−1 (gN −1 ) The columns of A∗ are the rows of A transposed (or written as columns) and conjugated. So column ` has χ`−1 (g0 ) χ (g ) `−1 1 .. . χ`−1 (gN −1 ) When you multiply AA∗ you multiply row j of A times column ` of A∗ to get the (j, `) entry of AA∗ . That is, the (j, `) entry of AA∗ is the inner product of vectors in CN N −1 X χj−1 (gk−1 )χ`−1 (gk−1 ) = hχj−1 , χ`−1 i k=0 and equals the inner product of the characters. We know hχj−1 , χ`−1 i = 0 if j 6= ` and hχj−1 , χj−1 i = N . So AA∗ has diagonal entries equal to N and off-diagonal entries 0, or AA∗ = N IN , as required. What is A∗ A? [Hint: what is A−1 ?] Solution: We have A(A∗ /N ) = IN and both A and A∗ /N are N ×N . This A−1 = (A∗ /N ) and (A∗ /N )A = A−1 A = IN . So A∗ A = N IN . If f : G → C is a function and vf is the N -tuple (f (g0 ), f (g1 ), . . . , f (gN −1 )) expressed as a column matrix, express the values fˆ(χk ) of the Fourier transform in terms of the matrix product Avf . This question should have said Āvf . Solution: The product Avf is N × 1 and has j th entry the product of row j of A times vf , that is N −1 N −1 X X χj−1 (gk−1 )f (gk−1 ) = f (gk−1 )χj−1 (gk−1 ) k=0 k=0 But, by definition 1 X f (g)χj−1 (g) fˆ(χj−1 ) = √ N g∈G N −1 1 X = √ f (gk−1 )χj−1 (gk−1 ) N k=0 N −1 1 X = √ χj−1 (gk−1 )f (gk−1 ) N k=0 So the column made up of the values fˆ(χj−1 ) is fˆ(χ0 ) f (g0 ) f (g1 ) fˆ(χ ) 1 1 1 √ √ = Āv = Ā . f . . .. . N N f (gN −1 ) fˆ(χN −1 ) Note: It might be better to take the row vft , the transpose of vf (or the values of f as a row) and the values of fˆ as a row. Then, transposing what we have above fˆ(χ0 ) fˆ(χ1 ) · · · 1 fˆ(χN −1 ) = √ vft A∗ N √ √ √ Alternatively since Ā/ N has inverse At / N . we could multiply what we got by At / N to get fˆ(χ0 ) fˆ(χ ) 1 1 √ At = vf . .. N fˆ(χN −1 ) Richard M. Timoney 2