Is it possible to find a matrix $A$ such that: $\exp(A)+\exp(A^{-1})=H_2$ with $A$ a $2\times 2$ matrix and $H_2$ a Hadamard matrix? The result can be extended to every Hadamard matrix $H_N$ with $N$ power of two? Thanks in advance.
A question about Hadamard matrices
-
0@Joriki:$A$can be complex. – 2012-10-09
1 Answers
Let $D=\left( \matrix {\sqrt{2}& 0\\0& -\sqrt{2}} \right)$ ,
$H_2=PD P^{-1}$, where $P=\left( \matrix {1+\sqrt{2}& 1-\sqrt{2}\\1& 1} \right)$.
$x=\frac{i\pi}{4} \implies e^x+e^{-x}=\sqrt{2}$
$x=\frac{3i\pi}{4} \implies e^x+e^{-x}=-\sqrt{2}$.
Let $E=\left( \matrix {\frac{i\pi}{4}& 0\\0&\frac{3i\pi}{4} } \right) $,
then $A=PEP^{-1}$ is such that $H_2=e^A+e^{-A}$.
$H_{2^n}=(\underbrace{P \otimes P\otimes ...\otimes P}_n)(\underbrace{D \otimes D\otimes ...\otimes D}_n)(\underbrace{P^{-1} \otimes...\otimes P^{-1}}_n)$, because $H_{2^n}=\underbrace{H \otimes ... \otimes H}_{n}$.
$\underbrace {D \otimes ... \otimes D}_n$ is a diagonal matrix.
For all $\lambda \in \mathbb{R}^*$, $x$ exists such that $e^x+e^{-x}=\lambda$.
Let $F$, a diagonal matrix, such that $e^F+e^{-F}=D \otimes ... \otimes D$, then $C=(P \otimes P\otimes ...\otimes P)F(P^{-1} \otimes...\otimes P^{-1})$ is such that $e^C+e^{-C}=H_{2^n}$
-
0And that is why I thought the problem was rather … strange. – 2012-10-09