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Let $H$ be a $d$ by $d$ real Hadamard matrix, namely: $$HH^{T}=d I$$ where $I$ is the identity matrix and $d=2^{k}$ for some natural number $k\geq 2$. The entries of $H$ are either $1$ or $-1$ and it has orthogonal rows. Can we say that this matrix has four $\frac{d}{2}$ by $\frac{d}{2}$ sub-matrices each of which have orthogonal rows?

If not, what conditions do we need on the matrix so that it would have such (Hadamard like) sub-matrices?

James Smithson
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    No. There exists a $12\times12$ Hadamard matrix, but there does not exist a $6\times6$ Hadamard matrix (because $4\nmid6$), and hence no submatrices of the type you described. – Jyrki Lahtonen Jun 16 '15 at 17:46
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    See Neil Sloane's page for more about Hadamard matrices. Link now corrected. – Jyrki Lahtonen Jun 16 '15 at 17:48
  • Thank you. Does this not contradict Sylvester construction of Hadamard matrices of order $2n$ using lower order Hadamard matrices or order $n$? – James Smithson Jun 16 '15 at 18:07
  • No. That construction shows that if there exists a Hadamard matrix of order $n$, then there also is one of order $2n$. It doesn't say anything about the converse. – Jyrki Lahtonen Jun 16 '15 at 18:08
  • I know and my question is under what conditions does the converse hold. If we say under no conditions the converse is true, which seemed to me to be what you are saying, then it means that I cannot construct a Hadamard matrix using Sylvester construction. Cuz surely if I construct a Hadamard matrix using Sylvester's method, then I have a Hadamard matrix that has four $\frac{d}{2}$ by $\frac{d}{2}$ Hadamard sub-matrices. No? – James Smithson Jun 16 '15 at 18:45
  • My point was simply to give a counterexample where the converse fails. The same applies to all known Hadamard matrices of order that is an odd multiple of four. I don't know about the converse in general. Sorry about not making that clear. – Jyrki Lahtonen Jun 16 '15 at 19:54
  • Thank you, I shall modify the question accordingly. – James Smithson Jun 16 '15 at 20:59
  • 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. – Will Orrick Jun 25 '15 at 15:53
  • @WillOrrick Thank you for your response. I'm mainly interested in solving this question: http://math.stackexchange.com/questions/1330576/determinant-of-a-certain-block-structured-positive-definite-matrix As you can see the eigenvalues of the Gram matrix $\lambda_{i}^{\pm}(\Gamma)=1\pm\sigma_{j}(B)$ where $\sigma_{j}(B)$ are the singular values of $B$. $B$ here is a Hadamard sub-matrix and I was hoping I could find cases where one can lower bound the minimum eigenvalue of the Gram matrix given a structure on $B$. – James Smithson Jun 26 '15 at 10:23
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    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. – Will Orrick Jul 01 '15 at 17:19
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    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? – Will Orrick Jul 01 '15 at 17:21
  • @WillOrrick Think of two $d$-dimensional ONB's that are mutually unbiased with respect to one another. Now choose $\frac{d}{2}$ elements from the first basis and $\frac{d}{2}$ elements from the second. Let's say the first $\frac{d}{2}$ elements from the two bases and put them in the vector set $V$. The Gram matrix assigned to $V$ is $\Gamma$. The off-diagonal blocks are therefore $\frac{1}{\sqrt{d}}S$ where $S$ is the sub-matrix of some Hadamard matrix. – James Smithson Jul 01 '15 at 21:10
  • @JyrkiLahtonen do you know if something can be said about the rank of $2^d$ by $2^d$ submatrices of a $2^{(d+1)}$ Walsh matrix? – gen Oct 08 '21 at 13:29
  • @gen No idea. Will Orrick seems to know more about this topic, and if he cannot say anything definitive, I take it as a sign that the question may be difficult in general. – Jyrki Lahtonen Oct 09 '21 at 12:06

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