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As shown in this answer, if $A$ is a real skew-symmetric matrix, and $v,w$ are a pair of orthogonal singular vectors with $$Av=sw \qquad\text{ and }\qquad Aw=-sv,$$ for some $s>0$, then the corresponding eigenvectors of $A$ are $v\pm iw$ (with eigenvalues $\pm is$).

However, as discussed for example in these notes (Link to pdf), one can state more generally that for any (possibly complex) skew-symmetric $A$ there is a unitary $U$ such that $U^TAU$ is block diagonal:

$$U^T A U = \operatorname{diag}\left\{\begin{pmatrix}0 & m_1 \\ -m_1 & 0\end{pmatrix}, \begin{pmatrix}0 & m_2 \\ -m_2 & 0\end{pmatrix},..., \begin{pmatrix}0 & m_n \\ -m_n & 0\end{pmatrix},\boldsymbol0\right\}.$$ In analogy with the above statement about the real case, this can be seen as equivalent to stating that the singular vectors of $A$ can be organised as pairs $v_k,w_k$ such that $$Av_k=s_kw_k \qquad\text{ and }\qquad Aw_k^*=-s_kv_k^*.$$ However, now $v_k\pm iw_k$ aren't always eigenvectors.

Is there a general relation between eigenvectors and singular vectors in the complex case?

glS
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1 Answers1

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There cannot be any such relation for generic complex matrices, because a skew-symmetric complex matrice need not be diagonalisable. For example,

$$\begin{pmatrix}0 & 1 & 0 \\ -1 & 0 & -i \\ 0 & i & 0 \end{pmatrix}$$ is not diagonalisable. Its singular values still pair as per the mentioned result: they are $0$ and $\pm\sqrt2$; however, the $\pm\sqrt2$ do not correspond to any eigenvector.

glS
  • 7,963
  • This matrix has characteristic polynomial of lambda^3 = 0, then how are we obtaining +sqrt(2), -sqrt(2), 0 as three values? – Diptangshu Feb 08 '25 at 07:07
  • @Diptangshu not sure what you mean. I'm not saying $\pm\sqrt2$ are eigenvalues. The only eigenvalue is indeed $0$, as per the characteristic polynomial. But the matrix is not diagonalisable, hence why it's possible that there aren't nonzero eigenvalues, but there are nonzero singular values – glS Feb 08 '25 at 08:38