Let $A\in\mathbb{R}^{n\times n}$ be a diagonalizable matrix with real and strictly positive eigenvalues (note that $A$ is not required to be symmetric). 



> 
> **My question.** Do there exist an orthogonal matrix $T\in\mathbb{R}^{n\times n}$ and a symmetric positive definite matrix $P\in\mathbb{R}^{n\times n}$ such that
>  $$
> TAPT^\top = D+S,
> $$
>  where $D\in\mathbb{R}^{n\times n}$ is a diagonal matrix with *positive* diagonal entries and $S\in\mathbb{R}^{n\times n}$ is a skew-symmetric matrix?
> 
> 
> 


Of course, if the diagonal entries of $D$ are not required to be positive then the answer is in the affirmative (see, e.g., [this related question](https://mathoverflow.net/questions/282694/is-every-real-matrix-conjugate-to-a-semi-antisymmetric-matrix)).



A "positive diagonal plus skew-symmetric" matrix decomposition
Choose any positive definite matrix $Q$. Since $A$ has eigenvalues with positive real part, the Lyapunov equation $$AP + PA^\top = Q$$
is solvable, and its solution $P$ is symmetric and positive definite.


Now decompose $AP = H+\hat{S}$, where $H$ is symmetric and $\hat{S}$ is skew-symmetric. Plugging this decomposition into the Lyapunov equation, we see that $H=\frac12 Q$. Then, take an eigendecomposition $\frac12 Q = TDT^\top$, with orthogonal $T$; since we chose $Q$ positive definite, $D$ has positive diagonal entries.


Hence we have
$$
AP = \frac12 Q + S = TDT^\top + \hat{S},
$$
with $\hat{S}$ skew-symmetric, or
$$
T^\top AP T = D + T^\top \hat{S} T,
$$
where $S=T^\top \hat{S} T$ is skew-symmetric.


