# A "positive diagonal plus skew-symmetric" matrix decomposition

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).

• Question 2 seems trivial - scaling $P$ scales $tr(D)$. Sep 25, 2018 at 17:24
• @user44191: Absolutely right! I will remove my second question right away. Sep 25, 2018 at 17:27

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.
Rewrite your equation: $$A = (T^{-1} (D+S) T) (T^{-1} P T)$$.
Choose $$B, C$$ symmetric positive definite with $$A = BC$$ (https://pure.tue.nl/ws/files/1810587/Metis198781.pdf ). Then choose $$D$$ similar to $$B$$, choose $$T$$ such that $$B = T^{-1} D T$$, and $$P = T C T^{-1}$$. Also choose $$S = 0$$.