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10h
comment Hadamard matrices and sub-matrices (Converse of Sylvester Construction)
By the way, for size 16, there are five equivalence classes, and all five have Hadamard submatrices of size 8. I have no idea what the situation is for size 32. I'm not sure whether I can be helpful to you. Can you say why you refer to $B$ as a Hadamard submatrix?
10h
comment Hadamard matrices and sub-matrices (Converse of Sylvester Construction)
The belief stated in my earlier comment - that it would be relatively easy to say something about Hadamard submatrices of Hadamard matrices whose size is an odd multiple of 8 - appears to have been unfounded. There are 60 equivalence classes of Hadamard matrices of size 24, and I expected that only a small subset of these with simple structure would have Hadamard submatrices of size 12. This is completely wrong. Now that I've done the calculation, I find that 58 of the 60 equivalence classes have such Hadamard submatrices, and that the structure can be more complicated than I thought.
1d
comment How to prove infinite solution vs no solution for singular matrix problem.
As for the general question: write your system as an augmented matrix and row-reduce. If the coefficient matrix is singular, you will end up with a all-zero row on the left. The system is consistent only if you also have 0 on the right.
1d
comment How to prove infinite solution vs no solution for singular matrix problem.
Should the $a$s in your definition of $B$ be $\alpha$s? Is your coefficient matrix written correctly? It doesn't look singular to me (except when $\alpha=0$).
Jun
25
comment Hadamard matrices and sub-matrices (Converse of Sylvester Construction)
I see that you changed to question in response to Jyrki Lahtonen's comments by restricting $d$ to powers of $2$ instead of letting it be any multiple of $4$. You could have eliminated his counterexample simply by requiring $d$ to be a multiple of $8$. My belief is that it's probably easier to say something about odd multiples of $8$ than about powers of $2$. In fact, I suspect that the higher the power of $2$ dividing $d$, there more difficult things become. If you are interested, I can try to elaborate.
Jun
24
comment Relation between number of unique values in Gramian Matrix (G) and the matrices that created it
I've been iteratively improving the answer, and now think I'm satisfied with it, unless there are questions. I plan to submit the sequence to OEIS.
Jun
24
revised Relation between number of unique values in Gramian Matrix (G) and the matrices that created it
nonrecursive formula
Jun
24
revised Relation between number of unique values in Gramian Matrix (G) and the matrices that created it
completed proof (sketch); removed extra equalities from deflation rules, leading to faster evaluation and greater compatibility with proof of the rules; modified examples accordingly
Jun
24
revised Relation between number of unique values in Gramian Matrix (G) and the matrices that created it
replaced proof of deflation rule 2 with a correct one; sketched proof of deflation rule 3
Jun
18
revised Relation between number of unique values in Gramian Matrix (G) and the matrices that created it
added 2463 characters in body
Jun
18
revised Relation between number of unique values in Gramian Matrix (G) and the matrices that created it
reformatted; corrected significant error in deflation rule 3; eliminated possibility of empty string as it is never used; started the proof
Jun
17
answered Relation between number of unique values in Gramian Matrix (G) and the matrices that created it
Jun
15
comment Visualising finite fields
The link seems no longer to work. Perhaps this is one of the papers you meant? If you know of other links that could be added, that would be very helpful.
Jun
13
comment Relation between number of unique values in Gramian Matrix (G) and the matrices that created it
I'd like to try to get your question reopened so I can post an answer. Before I do that, It might help if you would edit the question to include the definition of $M\times N$ Hadamard matrix that you spelled out in the comments. If you could say more about the Toeplitz question as well, that would help too. (What set are the elements taken from? What probability distribution are you assuming? Is there any reason to believe that, apart from the symmetry of $A'A$, the elements won't all be different?) Also, people on this site seem to like it when you explain how the question arose.
Jun
13
revised Why does the largest $x$ such that $a$, $b$ divided by $x$ leave the same remainder equal $a-b$?
added 3 characters in body
Jun
12
comment Obtain all combinations of 3 numbers with repetition.
If you need to carry out the process for a different $H$, and it isn't already row-reduced, you can use the Pari command matsolvemod(H,3,0) with flag=1 to obtain a set of basis vectors for the nullspace.
Jun
12
comment Obtain all combinations of 3 numbers with repetition.
I tried answering at s.overflow, but the post got deleted. The linear code is the nullspace of $H$. One can find a basis for the nullspace by row reduction, but in this example, the matrix is already reduced: columns 1, 2, and 5 contain pivots. So one can set the elements of the remaining 10 columns arbitrarily, and then compute the elements in positions 1, 2, and 5. In Pari you can generate all possible vectors by looping through the integers 0 through $3^{10}-1$. Use the command digits in base 3 to convert to a vector. (You'll have to pad with zeros.)
Jun
12
revised Why does the largest $x$ such that $a$, $b$ divided by $x$ leave the same remainder equal $a-b$?
edited title
Jun
12
comment Why does the largest $x$ such that $a$, $b$ divided by $x$ leave the same remainder equal $a-b$?
The integer quotient of $3$ by $96$ is $0$, with a remainder $3$. (Because $3=0\cdot96+3$.) Unless there is a compelling reason not to allow the quotient to be $0$, it simplifies the analysis considerably to do so.
Jun
12
comment Why does the largest $x$ such that $a$, $b$ divided by $x$ leave the same remainder equal $a-b$?
Thanks for the clarification and for editing the post. I still have a question however. You write $a>b>c>x$. Do you really mean to require that $x<c$? If so, then your claim is false, as a couple of posters have pointed out. Here's an example that shows the claim is false even for two numbers, $a$, $b$: let $a=105$, $b=9$. If $x$ is allowed to be greater than $9$, then $x=96=105-9$, and the remainder is $9$ for both $105$ and $9$. If $x$ must be less than $b$, then $x=6$, which is obviously not equal to $a-b$. The remainder in this case is $3$.