0
$\begingroup$

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, $U,\Sigma\to V$)?

$\endgroup$

1 Answer 1

0
$\begingroup$

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 $V = U\Omega$.

$\endgroup$
1
  • $\begingroup$ 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). $\endgroup$
    – Jellby
    Dec 4, 2021 at 11:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .