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$A$ is a symmetric matrix, i.e $A^*=A$. I already proved, that if $A$ is symmetric $\implies e^A$ is positive. Now I would like to know why every positive hermitian matrix B can be written as $e^A$.

Is it possible to prove it using the matrix logarithm?

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up vote 2 down vote accepted

Every positive Hermitian $N\times N$ matrix $B$ can be written as $B= U D U^\dagger$ with $U$ a unitary matrix and $$ D =\begin{pmatrix} d_1 \\ & \ddots\\ & & b_N \end{pmatrix} $$ with $d_j>0\quad \forall j$.

Now set $$A = U \begin{pmatrix} \log d_1 \\ & \ddots\\ & & \log b_N \end{pmatrix} U^\dagger$$ and it is easy to prove that $B= \exp A$.

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Thank you for your help. One more question: Is A uniquely defined ? – Montaigne Oct 29 '12 at 23:13
@user45170: Try adding $2\pi i$ to one of the $\log$s. – wj32 Oct 30 '12 at 8:05

Hermitian matrices are unitarily diagonalizable. Suppose we have $$A = UDU^{\dagger}$$ then $$\exp(A) = \sum_{n=1}^\infty\frac{1}{n!}UD^nU^\dagger=Ue^DU^\dagger$$ Clearly we have $$e^D = \rm{diag}(e^{\lambda_1},\ e^{\lambda_2},\ \cdots,\ e^{\lambda_n})$$ Reversing the above process will give us the desired result. Pick $\lambda_i$ so that each $e^{\lambda_i}$ is our desired eigenvalue (this is always positive since we assumed our Hermitian matrix to be positive - definite), we get that any Hermitian matrix is expressible as $e^A$ for some Hermitian $A$.

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