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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.

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I have no idea. But how does this problem arise? I am curious. –  Harald Hanche-Olsen Oct 9 '12 at 7:12
    
I found it on a paper of theoretical physics about superstrings. –  Riccardo.Alestra Oct 9 '12 at 7:18
1  
Hint: Try to use $A=a_0I+{\bf a}\cdot{\bf \sigma}$ with $\sigma_i$ Pauli matrices. –  Jon Oct 9 '12 at 7:35
    
Is $A$ supposed to be real or complex? –  joriki Oct 9 '12 at 8:17
    
@Joriki: $A$ can be complex. –  Riccardo.Alestra Oct 9 '12 at 9:43

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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 $e^C+e^{-C}=H_{2^n}$

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$A^{-1}$ isn't $-A$ –  Riccardo.Alestra Oct 9 '12 at 12:39
    
And that is why I thought the problem was rather … strange. –  Harald Hanche-Olsen Oct 9 '12 at 13:23

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