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If $A$ is an $n\times n$ integer matrix, then trivially $S=A+A^t$ and $P = AA^t$ where $t$ is ``transpose", are both symmetric.

Assume that $A$ is also a "$\lbrace -1,1 \rbrace$" matrix, i.e., the square of each entry in $A$ is equal to $1$.

Is there some rational-coefficient symmetric polynomial $P(x,y)$ (depending possibly on $A$ ?) such that $$ P(A,A^t) $$ is also a $\lbrace -1,1 \rbrace$ matrix?

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Luis, what is $A^*$? – Andrés E. Caicedo Jan 4 '11 at 1:26
I assume $A^t = A^*.$ – Igor Rivin Jan 4 '11 at 1:41
There are trivial solutions. E.g. $P(A, A^t) = I$. Or, if $P$ is allowed to depend on $A$, then choose $P_A(A, A^t)$ to be an arbitrary symmetric $\{-1,1\}$ matrix. Or to make it something less obvious, take the product of all characteristic polynomials of all such $n\times n$ matrices -- you don't need to use $A^t$, just $A$, and you'll get 0. Probably an interpolation could be used to make the function have more "interesting" values. You need a better criterion for this question to make the criterion clear --- perhaps the motivation would help. – Bill Thurston Jan 4 '11 at 3:35
@Bill: Re your last construction -- 0 is not actually $\pm 1,$ so that does not work, but I agree that allowing the polynomial to depend on $A$ is silly. Here is a more specific question (probably of no interest to the original poster): Is there a polynomial (possibly depending on $n$) so that $P(A, B) = A \circ B,$ where $\circ$ denotes the Hadamard (entriwise) product of $A$ and $B$ and $A, B$ are $n \times n.$ – Igor Rivin Jan 4 '11 at 4:47
@Igor, such $P(A,B)$ should have been $AB$ when restricted to the diagonal matrices, so it does not exist in general when $n\ge 2$. – Wadim Zudilin Jan 4 '11 at 6:52
up vote 5 down vote accepted

As already mentioned in the comments above, it's not a big deal to find such a polynomial for a particular $\pm1$ matrix. If the question is about a "universal" polynomial (that is, depending only on $n$), then I would expect the answer "no". For $n=2$, take the matrix $$ A=\left(\begin{matrix} 1 & -1 \cr 1 & 1 \end{matrix}\right). $$ Then $A+A^t=2I_2$ and $AA^t=2I_2$ where $I_2$ denotes the $2\times 2$ identity matrix. Because any symmetric polynomial with numerical (not matrix!) coefficients is a polynomial in $A+A^t$ and $AA^t$, its value will be always $cI_2$ for a certain numerical constant $c$. Therefore, the off-diagonal entries will be always zero.

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Your example generalizes (conjecturally) to all $n$ which are a multiple of $4$. It is conjectured that there is a skew Hadamard matrix of order $4k$ for each $k$, where a skew Hadamard matrix is a matrix $A$ with $\pm 1$ entries whose rows are orthogonal (so $AA^T=nI$) and $A+A^T=2I$. – Gjergji Zaimi Jan 4 '11 at 11:35
Thanks, Gjergji, for recalling this conjecture. I did not realize the connection... I wonder whether it is the reason for the OP. – Wadim Zudilin Jan 4 '11 at 12:19
There are counter-examples with circulant matrices ? – Luis H Gallardo Jan 8 '11 at 15:00

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