MA342A (Harmonic Analysis 1) Tutorial sheet 3
[October 29, 2015]
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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
0
