3
$\begingroup$

I've read that any real skew-symmetric matrix $A$, where $A^T = -A$, can be brought into block diagonal form

$ A = Q \, \Sigma \, Q^T = \left( \begin{array}{ccccl} \vec{q}_1 & \vec{q}_2 & \vec{q}_3 & \vec{q}_4 & \dots \end{array} \right) \, \left( \begin{array}{ccccl} 0 & \lambda_1 & 0 & 0 & \dots \\ -\lambda_1 & 0 & 0 & 0 & \dots \\ 0 & 0 & 0 & \lambda_2 & \dots \\ 0 & 0 & -\lambda_2 & 0 & \dots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array} \right) \, \left( \begin{array}{c} \vec{q}^T_1 \\ \vec{q}^T_2 \\ \vec{q}^T_3 \\ \vec{q}^T_4 \\ \vdots \end{array} \right)$

where $\vec{q}_i$ are real, orthogonal column vectors.

But also, the non-zero eigenvalues of $A$ are purely imaginary and occur in pairs $\pm i \lambda_i$, which are the same $\lambda_i$ as occur in the blocks of $\Sigma$. We could then also write $A$ in complex diagonal form

$ A = U \, D \, U^\dagger = \left( \begin{array}{ccccl} \vec{u}_1 & \vec{u}_2 & \vec{u}_3 & \vec{u}_4 & \dots \end{array} \right) \, \left( \begin{array}{ccccl} i\lambda_1 & 0 & 0 & 0 & \dots \\ 0 & -i\lambda_1 & 0 & 0 & \dots \\ 0 & 0 & i\lambda_2 & 0 & \dots \\ 0 & 0 & 0 & -i\lambda_2 & \dots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array} \right) \, \left( \begin{array}{c} \vec{u}^\dagger_1 \\ \vec{u}^\dagger_2 \\ \vec{u}^\dagger_3 \\ \vec{u}^\dagger_4 \\ \vdots \end{array} \right)$

where $\vec{u}_i$ are complex, orthogonal column vectors.

My question is what is the relationship between the complex eigenvectors $\vec{u}_i$ and the real vectors $\vec{q}_i$ that bring $A$ into block diagonal form? And how can I show the link?

$\endgroup$
6
$\begingroup$

Define $$ W = \frac 1{\sqrt 2}\pmatrix{1 & -i\\ i & -1} $$ note that $W$ is unitary (which neatly satisfies $W = W^\dagger$). Moreover, $$ \pmatrix{0 & \lambda\\-\lambda & 0} = W\pmatrix{\lambda i\\&-\lambda i}W^\dagger $$ We can then say that $\Sigma = \tilde WD\tilde W^\dagger$, where $\tilde W$ is the block-matrix $$ \tilde W = \pmatrix{W\\&W\\&&\ddots\\&&&W} = I_{n/2} \otimes W $$ Following your notation, we have $U = Q\tilde W$. It follows that $$ u_i = Q \tilde W e_i $$ where $e_i$ denotes the $i$th canonical basis vector. Going through the matrix multiplication, you'll find that $u_i$ can always be written as a linear combination of two vectors $v_j,v_{j+1}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy