Unbiased Bush-type Hadamard matricespeople.cst.cmich.edu/salis1bt/shrikhande/slides-kharaghani.pdf · Outline: I Hadamard matrix I The auxiliary matrices corresponding to a Hadamard

H. Kharaghani Unbiased Bush-type Hadamard matrices

Happy No Teaching Mohan!


Unbiased Bush-type Hadamard matrices

H. Kharaghani

University of LethbridgeCanada

Conference on Combinatorial Designs and Graph Theoryin honor of Mohan Shrikhande

on the occasion of his retirement

Department of MathematicsCentral Michigan University

October 10, 2015


Outline:

I Hadamard matrix

I The auxiliary matrices corresponding to a Hadamard matrix

I Bush-type Hadamard matrix

I Existence and examples

I Unbiased Hadamard matrices

I Association Schemes

I An upper bound for the number of unbiased Bush-type matrices

I The sharpness of the upper bound


Outline:

I Hadamard matrix









Outline:

I Hadamard matrix









Outline:

I Hadamard matrix









Outline:

I Hadamard matrix









Outline:

I Hadamard matrix









Outline:

I Hadamard matrix









Outline:

I Hadamard matrix









Outline:

I Hadamard matrix









DefinitionA square ±1-matrix that has pairwise orthogonal rows is called aHadamard matrix.

Example:

H2 =

(1 11 −

).

In order to have three (−1,1)-vectors of dimension n mutuallyorthogonal, it is necessary to have n = 4k .

Example: H4 = H2⊗H2 =

(H2 H2

H2 −H2

).

Hadamard conjecture: There exists a Hadamard matrix for anyorder 4n.

This conjecture is verified for all n < 167.



Example:

H2 =

(1 11 −

).



(H2 H2

H2 −H2

).





Example:

H2 =

(1 11 −

).



(H2 H2

H2 −H2

).





Example:

H2 =

(1 11 −

).

In order to have three (−1,1)-vectors of dimension n mutuallyorthogonal,

it is necessary to have n = 4k .


(H2 H2

H2 −H2

).





Example:

H2 =

(1 11 −

).



(H2 H2

H2 −H2

).





Example:

H2 =

(1 11 −

).



(H2 H2

H2 −H2

).





Example:

H2 =

(1 11 −

).



(H2 H2

H2 −H2

).





Example:

H2 =

(1 11 −

).



(H2 H2

H2 −H2

).




A characterization of Hadamard matrices

There is a Hadamard matrix H of order 2n if and only if there are 2n(−1,1) matrices C0,C1,C2, . . . ,C2n−1 of order 2n such that:

I Cti = Ci

I CiCj = 0, i 6= j

I C2i = 2nCi

I C0 +C1 +C2 + . . .+C2n−1 = 2nI2n

I C0 may be chosen to be the matrix of all ones.

Proof:Let ri be the i +1-th row of H and let Ci = r t

i ri , i = 0,1, · · · ,2n−1.




I Cti = Ci

I CiCj = 0, i 6= j

I C2i = 2nCi

I C0 +C1 +C2 + . . .+C2n−1 = 2nI2n



i ri , i = 0,1, · · · ,2n−1.




I Cti = Ci

I CiCj = 0, i 6= j

I C2i = 2nCi

I C0 +C1 +C2 + . . .+C2n−1 = 2nI2n



i ri , i = 0,1, · · · ,2n−1.




I Cti = Ci

I CiCj = 0, i 6= j

I C2i = 2nCi

I C0 +C1 +C2 + . . .+C2n−1 = 2nI2n



i ri , i = 0,1, · · · ,2n−1.




I Cti = Ci

I CiCj = 0, i 6= j

I C2i = 2nCi

I C0 +C1 +C2 + . . .+C2n−1 = 2nI2n



i ri , i = 0,1, · · · ,2n−1.




I Cti = Ci

I CiCj = 0, i 6= j

I C2i = 2nCi

I C0 +C1 +C2 + . . .+C2n−1 = 2nI2n



i ri , i = 0,1, · · · ,2n−1.




I Cti = Ci

I CiCj = 0, i 6= j

I C2i = 2nCi

I C0 +C1 +C2 + . . .+C2n−1 = 2nI2n



i ri , i = 0,1, · · · ,2n−1.




I Cti = Ci

I CiCj = 0, i 6= j

I C2i = 2nCi

I C0 +C1 +C2 + . . .+C2n−1 = 2nI2n



i ri , i = 0,1, · · · ,2n−1.




I Cti = Ci

I CiCj = 0, i 6= j

I C2i = 2nCi

I C0 +C1 +C2 + . . .+C2n−1 = 2nI2n



i ri , i = 0,1, · · · ,2n−1.


Bush–type Hadamard Matrices

A Bush-type Hadamard matrix is a block matrix H = [Hij ] of order 4n2

with block size 2n, Hii = J2n and HijJ2n = J2nHij = 0, i 6= j , 1≤ i ≤ 2n,1≤ j ≤ 2n, where J2n is the 2n by 2n matrix of all entries 1.

Example:

H16 =

1 1 1 1 1 1 − − 1 − 1 − 1 − − 11 1 1 1 1 1 − − − 1 − 1 − 1 1 −1 1 1 1 − − 1 1 1 − 1 − − 1 1 −1 1 1 1 − − 1 1 − 1 − 1 1 − − 11 1 − − 1 1 1 1 1 − − 1 1 − 1 −1 1 − − 1 1 1 1 − 1 1 − − 1 − 1− − 1 1 1 1 1 1 − 1 1 − 1 − 1 −− − 1 1 1 1 1 1 1 − − 1 − 1 − 11 − 1 − 1 − − 1 1 1 1 1 1 1 − −− 1 − 1 − 1 1 − 1 1 1 1 1 1 − −1 − 1 − − 1 1 − 1 1 1 1 − − 1 1− 1 − 1 1 − − 1 1 1 1 1 − − 1 11 − − 1 1 − 1 − 1 1 − − 1 1 1 1− 1 1 − − 1 − 1 1 1 − − 1 1 1 1− 1 1 − 1 − 1 − − − 1 1 1 1 1 11 − − 1 − 1 − 1 − − 1 1 1 1 1 1




with block size 2n,

Hii = J2n and HijJ2n = J2nHij = 0, i 6= j , 1≤ i ≤ 2n,1≤ j ≤ 2n, where J2n is the 2n by 2n matrix of all entries 1.

Example:

H16 =

1 1 1 1 1 1 − − 1 − 1 − 1 − − 11 1 1 1 1 1 − − − 1 − 1 − 1 1 −1 1 1 1 − − 1 1 1 − 1 − − 1 1 −1 1 1 1 − − 1 1 − 1 − 1 1 − − 11 1 − − 1 1 1 1 1 − − 1 1 − 1 −1 1 − − 1 1 1 1 − 1 1 − − 1 − 1− − 1 1 1 1 1 1 − 1 1 − 1 − 1 −− − 1 1 1 1 1 1 1 − − 1 − 1 − 11 − 1 − 1 − − 1 1 1 1 1 1 1 − −− 1 − 1 − 1 1 − 1 1 1 1 1 1 − −1 − 1 − − 1 1 − 1 1 1 1 − − 1 1− 1 − 1 1 − − 1 1 1 1 1 − − 1 11 − − 1 1 − 1 − 1 1 − − 1 1 1 1− 1 1 − − 1 − 1 1 1 − − 1 1 1 1− 1 1 − 1 − 1 − − − 1 1 1 1 1 11 − − 1 − 1 − 1 − − 1 1 1 1 1 1




with block size 2n, Hii = J2n

and HijJ2n = J2nHij = 0, i 6= j , 1≤ i ≤ 2n,1≤ j ≤ 2n, where J2n is the 2n by 2n matrix of all entries 1.

Example:

H16 =

1 1 1 1 1 1 − − 1 − 1 − 1 − − 11 1 1 1 1 1 − − − 1 − 1 − 1 1 −1 1 1 1 − − 1 1 1 − 1 − − 1 1 −1 1 1 1 − − 1 1 − 1 − 1 1 − − 11 1 − − 1 1 1 1 1 − − 1 1 − 1 −1 1 − − 1 1 1 1 − 1 1 − − 1 − 1− − 1 1 1 1 1 1 − 1 1 − 1 − 1 −− − 1 1 1 1 1 1 1 − − 1 − 1 − 11 − 1 − 1 − − 1 1 1 1 1 1 1 − −− 1 − 1 − 1 1 − 1 1 1 1 1 1 − −1 − 1 − − 1 1 − 1 1 1 1 − − 1 1− 1 − 1 1 − − 1 1 1 1 1 − − 1 11 − − 1 1 − 1 − 1 1 − − 1 1 1 1− 1 1 − − 1 − 1 1 1 − − 1 1 1 1− 1 1 − 1 − 1 − − − 1 1 1 1 1 11 − − 1 − 1 − 1 − − 1 1 1 1 1 1





Example:

H16 =

1 1 1 1 1 1 − − 1 − 1 − 1 − − 11 1 1 1 1 1 − − − 1 − 1 − 1 1 −1 1 1 1 − − 1 1 1 − 1 − − 1 1 −1 1 1 1 − − 1 1 − 1 − 1 1 − − 11 1 − − 1 1 1 1 1 − − 1 1 − 1 −1 1 − − 1 1 1 1 − 1 1 − − 1 − 1− − 1 1 1 1 1 1 − 1 1 − 1 − 1 −− − 1 1 1 1 1 1 1 − − 1 − 1 − 11 − 1 − 1 − − 1 1 1 1 1 1 1 − −− 1 − 1 − 1 1 − 1 1 1 1 1 1 − −1 − 1 − − 1 1 − 1 1 1 1 − − 1 1− 1 − 1 1 − − 1 1 1 1 1 − − 1 11 − − 1 1 − 1 − 1 1 − − 1 1 1 1− 1 1 − − 1 − 1 1 1 − − 1 1 1 1− 1 1 − 1 − 1 − − − 1 1 1 1 1 11 − − 1 − 1 − 1 − − 1 1 1 1 1 1





Example:

H16 =

1 1 1 1 1 1 − − 1 − 1 − 1 − − 11 1 1 1 1 1 − − − 1 − 1 − 1 1 −1 1 1 1 − − 1 1 1 − 1 − − 1 1 −1 1 1 1 − − 1 1 − 1 − 1 1 − − 11 1 − − 1 1 1 1 1 − − 1 1 − 1 −1 1 − − 1 1 1 1 − 1 1 − − 1 − 1− − 1 1 1 1 1 1 − 1 1 − 1 − 1 −− − 1 1 1 1 1 1 1 − − 1 − 1 − 11 − 1 − 1 − − 1 1 1 1 1 1 1 − −− 1 − 1 − 1 1 − 1 1 1 1 1 1 − −1 − 1 − − 1 1 − 1 1 1 1 − − 1 1− 1 − 1 1 − − 1 1 1 1 1 − − 1 11 − − 1 1 − 1 − 1 1 − − 1 1 1 1− 1 1 − − 1 − 1 1 1 − − 1 1 1 1− 1 1 − 1 − 1 − − − 1 1 1 1 1 11 − − 1 − 1 − 1 − − 1 1 1 1 1 1





Example:

H16 =

1 1 1 1 1 1 − − 1 − 1 − 1 − − 11 1 1 1 1 1 − − − 1 − 1 − 1 1 −1 1 1 1 − − 1 1 1 − 1 − − 1 1 −1 1 1 1 − − 1 1 − 1 − 1 1 − − 11 1 − − 1 1 1 1 1 − − 1 1 − 1 −1 1 − − 1 1 1 1 − 1 1 − − 1 − 1− − 1 1 1 1 1 1 − 1 1 − 1 − 1 −− − 1 1 1 1 1 1 1 − − 1 − 1 − 11 − 1 − 1 − − 1 1 1 1 1 1 1 − −− 1 − 1 − 1 1 − 1 1 1 1 1 1 − −1 − 1 − − 1 1 − 1 1 1 1 − − 1 1− 1 − 1 1 − − 1 1 1 1 1 − − 1 11 − − 1 1 − 1 − 1 1 − − 1 1 1 1− 1 1 − − 1 − 1 1 1 − − 1 1 1 1− 1 1 − 1 − 1 − − − 1 1 1 1 1 11 − − 1 − 1 − 1 − − 1 1 1 1 1 1


The existence of Bush-type Hadamard matrices

Bush[1972]There is a symmetric Bush-type Hadamard matrix of order 4n2for eachn for which there is a projective plane of order 2n.

There is a Bush-type Hadamard matrix of order 4n2 for each positiveinteger n for which there is a Hadamard matrix of order 2n.

Muzychuck, Xinag [2006]There is a symmetric Bush-type Hadamard matrix of order 4n4 foreach odd integer n.

The first order for which the existence of a Bush-type Hadamard matrixis not known is 4(72) = 196.



Bush[1972]

There is a symmetric Bush-type Hadamard matrix of order 4n2for eachn for which there is a projective plane of order 2n.






Bush[1972]There is a symmetric Bush-type Hadamard matrix of order 4n2

for eachn for which there is a projective plane of order 2n.













There is a Bush-type Hadamard matrix of order 4n2

for each positiveinteger n for which there is a Hadamard matrix of order 2n.













Muzychuck, Xinag [2006]

There is a symmetric Bush-type Hadamard matrix of order 4n4 foreach odd integer n.















Unbiasedness

Let’s start with an example of unbiasedness:Let

H =

1 1 1 11 1 − −1 − 1 −1 − − 1

,

K =

1 1 1 −1 1 − 11 − 1 1− 1 1 1

.

Then

HK t = 2

1 1 1 11 1 − −1 − 1 −− 1 1 −

.

The pair of Hadamard matrices H and K is an example of twounbiased Hadamard matrices of order 4. Hadamard matrices H,K oforder n are called unbiased if all the entries of HK t are of the sameabsolute values (which is neccesarily

√n).


Unbiasedness

Let’s start with an example of unbiasedness:

Let

H =

1 1 1 11 1 − −1 − 1 −1 − − 1

,

K =

1 1 1 −1 1 − 11 − 1 1− 1 1 1

.

Then

HK t = 2

1 1 1 11 1 − −1 − 1 −− 1 1 −

.


√n).


Unbiasedness


H =

1 1 1 11 1 − −1 − 1 −1 − − 1

,

K =

1 1 1 −1 1 − 11 − 1 1− 1 1 1

.

Then

HK t = 2

1 1 1 11 1 − −1 − 1 −− 1 1 −

.


√n).


Unbiasedness


H =

1 1 1 11 1 − −1 − 1 −1 − − 1

,

K =

1 1 1 −1 1 − 11 − 1 1− 1 1 1

.

Then

HK t = 2

1 1 1 11 1 − −1 − 1 −− 1 1 −

.


√n).


Unbiasedness


H =

1 1 1 11 1 − −1 − 1 −1 − − 1

,

K =

1 1 1 −1 1 − 11 − 1 1− 1 1 1

.

Then

HK t = 2

1 1 1 11 1 − −1 − 1 −− 1 1 −

.


√n).


Unbiasedness


H =

1 1 1 11 1 − −1 − 1 −1 − − 1

,

K =

1 1 1 −1 1 − 11 − 1 1− 1 1 1

.

Then

HK t = 2

1 1 1 11 1 − −1 − 1 −− 1 1 −

.

The pair of Hadamard matrices H and K is an example of twounbiased Hadamard matrices of order 4.

Hadamard matrices H,K oforder n are called unbiased if all the entries of HK t are of the sameabsolute values (which is neccesarily

√n).


Unbiasedness


H =

1 1 1 11 1 − −1 − 1 −1 − − 1

,

K =

1 1 1 −1 1 − 11 − 1 1− 1 1 1

.

Then

HK t = 2

1 1 1 11 1 − −1 − 1 −− 1 1 −

.


√n).


An upper bound

A set H of Hadamard matrices of order n is called mutually unbiased,if every distict pair of entries in H is unbiased.

I Let m be the number of elements in a set of mutually unbiasedHadamard matrices of order 4n2, n odd, then m = 2. The firstopen order is 100.

I Let m be the number of elements in a set of mutually unbiasedHadamard matrices of order 4n2, then m ≤ 2n2. The inequality issharp (Kerdock codes).

The situation for unbiased Bush-type Hadamard matrices is quitedifferent.Theorem [K,Sasani, Suda 2015]: Let m be the number of mutuallyunbiased Bush-type Hadamard matrices of order 4n2, thenm ≤ 2n−1. The inequality is sharp.


An upper bound






An upper bound


I Let m be the number of elements in a set of mutually unbiasedHadamard matrices of order 4n2, n odd, then m = 2.

The firstopen order is 100.




An upper bound






An upper bound






An upper bound




The situation for unbiased Bush-type Hadamard matrices is quitedifferent.

Theorem [K,Sasani, Suda 2015]: Let m be the number of mutuallyunbiased Bush-type Hadamard matrices of order 4n2, thenm ≤ 2n−1. The inequality is sharp.


An upper bound




The situation for unbiased Bush-type Hadamard matrices is quitedifferent.Theorem [K,Sasani, Suda 2015]: Let m be the number of mutuallyunbiased Bush-type Hadamard matrices of order 4n2, thenm ≤ 2n−1.

The inequality is sharp.


An upper bound






Some basic definitions and results

Lemma: Let H and K be two unbiased Bush-type Hadamard matricesof order 4n2. Let L be a (1,−1)-matrix so that HK t = 2nL. Then L is aBush-type Hadamard matrix.

Two Latin squares L1 and L2 of size n on symbol set {0,1,2, . . . ,n−1}are called suitable if every superimposition of each row of L1 on eachrow of L2 results in only one element of the form (a,a).Example:

L1 =

0 1 22 0 11 2 0

,

L2 =

0 2 11 0 22 1 0

,

are examples of two suitable Latin Squares on {0,1,2}.There are m mutually suitable Latin Squares if and only if there are mMutually Orthogonal Latin Squares (MOLS).



Lemma:

Let H and K be two unbiased Bush-type Hadamard matricesof order 4n2. Let L be a (1,−1)-matrix so that HK t = 2nL. Then L is aBush-type Hadamard matrix.


L1 =

0 1 22 0 11 2 0

,

L2 =

0 2 11 0 22 1 0

,




Lemma: Let H and K be two unbiased Bush-type Hadamard matricesof order 4n2.

Let L be a (1,−1)-matrix so that HK t = 2nL. Then L is aBush-type Hadamard matrix.


L1 =

0 1 22 0 11 2 0

,

L2 =

0 2 11 0 22 1 0

,






L1 =

0 1 22 0 11 2 0

,

L2 =

0 2 11 0 22 1 0

,





Two Latin squares L1 and L2 of size n on symbol set {0,1,2, . . . ,n−1}are called suitable

if every superimposition of each row of L1 on eachrow of L2 results in only one element of the form (a,a).Example:

L1 =

0 1 22 0 11 2 0

,

L2 =

0 2 11 0 22 1 0

,





Two Latin squares L1 and L2 of size n on symbol set {0,1,2, . . . ,n−1}are called suitable if every superimposition of each row of L1 on eachrow of L2 results in only one element of the form (a,a).

Example:L1 =

0 1 22 0 11 2 0

,

L2 =

0 2 11 0 22 1 0

,






L1 =

0 1 22 0 11 2 0

,

L2 =

0 2 11 0 22 1 0

,

are examples of two suitable Latin Squares on {0,1,2}.

There are m mutually suitable Latin Squares if and only if there are mMutually Orthogonal Latin Squares (MOLS).





L1 =

0 1 22 0 11 2 0

,

L2 =

0 2 11 0 22 1 0

,






L1 =

0 1 22 0 11 2 0

,

L2 =

0 2 11 0 22 1 0

,



A lower bound

If there are m mutually suitable Latin squares of size 2n on{0,1,2 . . . ,2n−1}with all zero entries on diagonal and a Hadamardmatrix of order 2n, then there are m mutually unbiased Bush-typeHadamard matrices of order 4n2.Proof: Use the auxiliary matrices constructed from the Hadamardmatrix of order 2n in the m mutually suitable Latin squares of size 2non {0,1,2 . . . ,2n−1}.

I Cti = Ci

I CiCj = 0, i 6= j

I C2i = 2nCi

I C0 +C1 +C2 + . . .+C2n−1 = 2nI2n



A lower bound

If there are m mutually suitable Latin squares of size 2n on{0,1,2 . . . ,2n−1}

with all zero entries on diagonal and a Hadamardmatrix of order 2n, then there are m mutually unbiased Bush-typeHadamard matrices of order 4n2.Proof: Use the auxiliary matrices constructed from the Hadamardmatrix of order 2n in the m mutually suitable Latin squares of size 2non {0,1,2 . . . ,2n−1}.

I Cti = Ci

I CiCj = 0, i 6= j

I C2i = 2nCi

I C0 +C1 +C2 + . . .+C2n−1 = 2nI2n



A lower bound

If there are m mutually suitable Latin squares of size 2n on{0,1,2 . . . ,2n−1}with all zero entries on diagonal and a Hadamardmatrix of order 2n,

then there are m mutually unbiased Bush-typeHadamard matrices of order 4n2.Proof: Use the auxiliary matrices constructed from the Hadamardmatrix of order 2n in the m mutually suitable Latin squares of size 2non {0,1,2 . . . ,2n−1}.

I Cti = Ci

I CiCj = 0, i 6= j

I C2i = 2nCi

I C0 +C1 +C2 + . . .+C2n−1 = 2nI2n



A lower bound

If there are m mutually suitable Latin squares of size 2n on{0,1,2 . . . ,2n−1}with all zero entries on diagonal and a Hadamardmatrix of order 2n, then there are m mutually unbiased Bush-typeHadamard matrices of order 4n2.

Proof: Use the auxiliary matrices constructed from the Hadamardmatrix of order 2n in the m mutually suitable Latin squares of size 2non {0,1,2 . . . ,2n−1}.

I Cti = Ci

I CiCj = 0, i 6= j

I C2i = 2nCi

I C0 +C1 +C2 + . . .+C2n−1 = 2nI2n



A lower bound


I Cti = Ci

I CiCj = 0, i 6= j

I C2i = 2nCi

I C0 +C1 +C2 + . . .+C2n−1 = 2nI2n



A lower bound


I Cti = Ci

I CiCj = 0, i 6= j

I C2i = 2nCi

I C0 +C1 +C2 + . . .+C2n−1 = 2nI2n



A lower bound

Example:

L1 =

0 1 2 31 0 3 22 3 0 13 2 1 0

,

L2 =

0 3 1 23 0 2 11 2 0 32 1 3 0

,

are two suitable Latin Squares on {0,1,2,3}. Replace i by Ci

constructed from a Hadamard matrix of order 4.

Theorem [Holzmann, K, Orrick 2009]: There are 4n−1 Mutuallyunbiased Bush-type Hadamard matrices of order 42n.


A lower bound

Example:

L1 =

0 1 2 31 0 3 22 3 0 13 2 1 0

,

L2 =

0 3 1 23 0 2 11 2 0 32 1 3 0

,

are two suitable Latin Squares on {0,1,2,3}.

Replace i by Ci




A lower bound

Example:

L1 =

0 1 2 31 0 3 22 3 0 13 2 1 0

,

L2 =

0 3 1 23 0 2 11 2 0 32 1 3 0

,





A lower bound

Example:

L1 =

0 1 2 31 0 3 22 3 0 13 2 1 0

,

L2 =

0 3 1 23 0 2 11 2 0 32 1 3 0

,





A lower bound

Example:

L1 =

0 1 2 31 0 3 22 3 0 13 2 1 0

,

L2 =

0 3 1 23 0 2 11 2 0 32 1 3 0

,





Association schemes

A symmetric d-class association scheme,

with vertex set X of size nand d classes is a set of symmetric (0,1)-matrices A0, . . . ,Ad , whichare not equal to zero matrix, with rows and columns indexed by X ,such that:

1. A0 = In, the identity matrix of order n.

2. ∑di=0 Ai = Jn, the matrix of order n with all one’s entries.

3. For all i , j , AiAj = ∑dk=0 pk

ij Ak for some pkij ’s.

The vector space spanned by Ai ’s forms a commutative algebra,denoted by A and called the Bose-Mesner algebra or adjacencyalgebra.


Association schemes

A symmetric d-class association scheme, with vertex set X of size nand d classes is a set of symmetric (0,1)-matrices A0, . . . ,Ad , whichare not equal to zero matrix,

with rows and columns indexed by X ,such that:







Association schemes

A symmetric d-class association scheme, with vertex set X of size nand d classes is a set of symmetric (0,1)-matrices A0, . . . ,Ad , whichare not equal to zero matrix, with rows and columns indexed by X ,such that:







Association schemes








Association schemes








Association schemes








Association schemes








There exists a basis for A consisting of primitive idempotents, sayE0 = (1/n)Jn,E1, . . . ,Ed .

the algebra A is closed under the entrywise

multiplication denoted by ◦. The Krein parameters q(k)ij ’s are defined

by Ei ◦Ej = 1n ∑

dk=0 q(k)

ij Ek .

Lemma: The Krein parameters q(k)ij of an association scheme are

non-negative.This leads to the following:

Theorem [K,Sasani, Suda 2015]: Let m be the number of mutuallyunbiased Bush-type Hadamard matrices of order 4n2, thenm ≤ 2n−1. The inequality is sharp. There are 4n−1 MUBH matrixof order 42n.


There exists a basis for A consisting of primitive idempotents, sayE0 = (1/n)Jn,E1, . . . ,Ed . the algebra A is closed under the entrywise

multiplication denoted by ◦.

The Krein parameters q(k)ij ’s are defined


dk=0 q(k)

ij Ek .








dk=0 q(k)

ij Ek .








dk=0 q(k)

ij Ek .


non-negative.

This leads to the following:






dk=0 q(k)

ij Ek .








dk=0 q(k)

ij Ek .



Theorem [K,Sasani, Suda 2015]: Let m be the number of mutuallyunbiased Bush-type Hadamard matrices of order 4n2, thenm ≤ 2n−1.

The inequality is sharp. There are 4n−1 MUBH matrixof order 42n.





dk=0 q(k)

ij Ek .



Theorem [K,Sasani, Suda 2015]: Let m be the number of mutuallyunbiased Bush-type Hadamard matrices of order 4n2, thenm ≤ 2n−1. The inequality is sharp.

There are 4n−1 MUBH matrixof order 42n.





dk=0 q(k)

ij Ek .





Thanks

Many thanks to the organizers!


Thanks



Thanks



Happy Retirement Mohan!


Happy Retirement Yury!


Documents

Unbiased Bush-type Hadamard matricespeople.cst.cmich.edu/salis1bt/shrikhande/slides-kharaghani.pdf · Outline: I Hadamard matrix I The auxiliary matrices corresponding to a Hadamard