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Show that an $n \times n$ real matrix $A$ is symmetric iff $A$ can be written as $$A=P-Q$$ where $P$ and $Q$ are some $n \times n$ positive definite matrices. Can there be anything said similarly about complex $n \times n$ matrices?

Attempt: Clearly if $A=P-Q$, it is symmetric. Conversely, suppose $A$ is symmetric. Then $$A=R^{-1}DR$$ where $D=$diag$\lbrace \lambda_1,...\lambda_n \rbrace, \lambda_i \in \mathrm{R}$ so $$A=R^{-1}D_1R - R^{-1}D_2R$$ where $D_1=$diag$\lbrace \alpha_1,...\alpha_n \rbrace$ and $D_2=$diag$\lbrace \mu_1,...\mu_n\rbrace$ where $$\lambda_i = \alpha_i-\mu_i \quad s.t. \quad \alpha_i,\mu_i >0$$ We know that the last statement is possible, since the eigenvalues $\lambda_i$ of $A$ are real numbers.

We cannot conclude the same about complex $n \times n$ matrices.

I am not sure if there is more to this than what I proved above.

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  • $\begingroup$ Did you mean to impose as well the condition that $P$ and $Q$ be symmetric? $\endgroup$ – Robert Lewis Aug 11 '13 at 8:13
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For the complex case, substitute "hermitian" for "symmetric"

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