# Symplectic positive definite matrix.

I want to prove that any symmetric positive definite symplectic matrix, $A$, and any real number $\alpha >0$, also $A^{\alpha} \in \operatorname{Sp}(2n)$.

I was given a hint to decompose $\mathbb{R}^{2n}$ into direct sum of $V_{\lambda}$ the eigenspaces, where $\lambda \in \sigma(A)$, and then use the characterization of the eigenvalues of $A$.

I must say, that I feel totally clueless here. :-(

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Let $v$ be some eigenvector of $A$ with eigenvalue $\lambda$. By assumption $\lambda$ is real and positive. Suppose that $w$ is some other eigenvector with eigenvalue $\mu$. Then $$\omega(v, w) = \omega(Av, Aw) = \omega(\lambda v, \mu w) = \lambda \mu \omega(v, w).$$
Hence, for all such pairs $(v,w)$, either $\omega(v,w) = 0$ or $\lambda\mu = 1$. Hence the eigenvalues come in pairs $(\lambda, \lambda^{-1})$ and we can find a symplectic basis $\{x_1, p_1, \cdots, x_n, p_n)$ so that $$A x_i = \lambda_i x_i ,$$ $$A p_i = \lambda_i^{-1} p_i.$$ Then $A^\alpha$ acts as $$A^\alpha x_i = \lambda_i^\alpha x_i ,$$ $$A^\alpha p_i = \lambda_i^{-\alpha} p_i,$$ and since this basis is symplectic it is obvious that $A^\alpha$ is symplectic.