Symmetric polynomials preserving $-1,1$ matrices - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T22:16:42Z http://mathoverflow.net/feeds/question/51079 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/51079/symmetric-polynomials-preserving-1-1-matrices Symmetric polynomials preserving $-1,1$ matrices Luis H Gallardo 2011-01-04T01:09:30Z 2011-01-04T10:05:56Z <p>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.</p> <p>Assume that $A$ is also a "$\lbrace -1,1 \rbrace$" matrix, i.e., the square of each entry in $A$ is equal to $1$.</p> <p>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?</p> http://mathoverflow.net/questions/51079/symmetric-polynomials-preserving-1-1-matrices/51102#51102 Answer by Wadim Zudilin for Symmetric polynomials preserving $-1,1$ matrices Wadim Zudilin 2011-01-04T10:05:56Z 2011-01-04T10:05:56Z <p>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 $$A=\left(\begin{matrix} 1 &amp; -1 \cr 1 &amp; 1 \end{matrix}\right).$$ 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.</p>