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I don't know how to solve (if it's possible) the following matrix equation: $$\exp(H)=H^2,$$ where $H$ is a $N \times N$ hermitian matrix. Does someone know if this equation has solutions and if the answer is yes, how to solve it? Thanks in advance.

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    $\begingroup$ By the spectral theorem, $H$ has an orthonormal basis of eigenvectors. Look at what each side does to an eigenvector of $H$. $\endgroup$ Jun 13 '12 at 9:55
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    $\begingroup$ Just considering the case $N = 1$, we can find that it is in general impossible to obtain an algebraic solution, unless we allow the use of the Lambert $W$-function. Thus in analogy we may define the solution to a prototypical equation as a new function and then use it to express a solution to your equation. Or, if you are only interested in the numerical value, you may use multivariate Newton-Rapson method. $\endgroup$ Jun 13 '12 at 9:57
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  • If you consider $P^{—1}HP$ where $P$ is invertible, and $H$ is a solution, then this matrix will be a solution. Since the matrix is supposed to be hermitian you can find an unitary matrix $P$ such that $P^*DP=H$, where $D$ is diagonal.
  • Hence we deal the case $H$ diagonal. Each diagonal element satisfies the equation $e^x=x^2$, which is difficult to solve (we have to use special functions).
  • The solutions are the matrices of the form $P^*DP$, where $P$ is unitary and $D=\operatorname{diag}(\lambda_1,\ldots,\lambda_N)$, with $\lambda_j^2=e^{\lambda_j}$ for all $j\in\{1,\ldots,N\}$.
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    $\begingroup$ Shouldn't $P$ be a unitary matrix? $\endgroup$
    – user856
    Jun 13 '12 at 10:08
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I would like to give a different route to prove the conclusion of Davide's answer. As already stated, $H$ has a complete set of eigenfunctions and eigenvalues due to spectral theorem. So, given that $Hu_n=\lambda_n u_n$, we can consider the identity matrix written down as $I=\sum_n u_nu_n^\dagger$ and so, the initial identity just becomes $$ \sum_n(e^{\lambda_n}-\lambda_n^2)u_nu_n^\dagger=0 $$ and the equation for the eigenvalues immediately follows $e^{\lambda_n}-\lambda^2_n=0$. The solutions are given through Lambert W-function as already stated.

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  • $\begingroup$ I think no one has mentioned that the solution is unique (for a given $N$). The spectral theorem makes the eigenvalues real, and there is only one real solution to $e^x = x^2$, so $H$ must be a multiple of the identity. $\endgroup$
    – Erick Wong
    Jun 13 '12 at 16:39
  • $\begingroup$ Nice comment, thanks. $\endgroup$
    – Jon
    Jun 13 '12 at 16:43

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