# Is there are relationship between the eigenvectors and the real Schur vectors of a real skew-symmetric matrix?

A real skew-symmetric matrix $$A$$ can be diagonalized with complex eigenvectors and pure imaginary eigenvalues:

$$A=V S V^*$$

where $$S$$ is:

$$S = \begin{pmatrix} -\lambda_1\mathrm{i} & 0 & 0 & 0 & \cdots & 0 & 0 \\ 0 & \lambda_1\mathrm{i} & 0 & 0 & \cdots & 0 & 0 \\ 0 & 0 & -\lambda_2\mathrm{i} & 0 & \cdots & 0 & 0 \\ 0 & 0 & 0 & \lambda_2\mathrm{i} & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 0 & 0 & \cdots & 0 & 0 \end{pmatrix}$$ all $$\lambda_i$$ are real and positive, and $$V$$ is a complex unitary matrix.

Similarly, $$A$$ can be real-Schur-decomposed, with both real Schur form and vectors, i.e.:

$$A = U \Sigma U^\mathrm{T}$$

with $$\Sigma$$ given by the same $$\lambda_i$$’s:

$$\Sigma = \begin{pmatrix} 0 & \lambda_1 & 0 & 0 & \cdots & 0 & 0 \\ -\lambda_1 & 0 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 0 & \lambda_2 & \cdots & 0 & 0 \\ 0 & 0 & -\lambda_2 & 0 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 0 & 0 & \cdots & 0 & 0 \end{pmatrix}$$ and $$U$$ a real unitary (orthogonal) matrix.

Is there any relationship between $$V$$ and $$U$$. Specifically, given $$V$$ (and $$S$$), is there an “easy” way to get $$U$$ (or vice-versa, $$U,\Sigma\to V$$)?

Yes there is. Denote $$D = \pmatrix{-i & 0\\ 0 & i}, \quad J = \pmatrix{0 & 1\\ -1 & 0}, \quad W = \frac 1{\sqrt{2}}\pmatrix{i & 1\\1 & i}.$$ We find that $$W^*JW = D$$. Consequently, if $$\Sigma = \pmatrix{\lambda_1 J \\ & \ddots \\ && \lambda_k J\\ &&& 0},$$ Then we have $$\Omega^* \Sigma \Omega = S$$, where $$\Omega = \pmatrix{W\\ & \ddots \\ && W\\ &&& I}.$$ Now, if $$A = VSV^*$$, then $$A = V\Omega^* \Sigma \Omega V^* = (V\Omega^*)\Sigma (V \Omega^*)^*$$. In other words, if we are given $$V$$, then $$U = V\Omega^*$$. Conversely, if we are given $$U$$, then $$V = U\Omega$$.
• Thanks. When the eigenvectors $V$ are given as conjugate pairs, the corresponding $U$ vectors are then easily obtained as $R_i+I_i$ and $R_i-I_i$, where $R_i$ and $I_i$ are the real and imaginary parts of each conjugate pair. What confused me is that apparently each pair of $U$ vectors can be further rotated by an arbitrary angle, so $U$ and $V\Omega^*$ obtained from independent decompositions are not necessarily equal (even after making sure the orders match). Dec 4, 2021 at 11:25