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I checked some examples and I always received that skew-symmetric matrix of even dimension has only pure imaginary eigenvalues.

For example:
$\begin{bmatrix} 0 & 2 & 3 & 1 \\ -2 & 0 & 1 & 4 \\ -3 & -1 & 0 & 1 \\ -1 & -4 & -1 & 0 \end{bmatrix}$

Eigenvalues: $( 0.000, 5.406i) ( 0.000,-5.406i) ( 0.000, 1.665i) ( 0.000,-1.665i)$

How can be explained such property?
Additionally why skew-symmetric of even dimension has non-zero determinant in opposition to odd dimensional skew-symmetric matrices ?

(I'm not considering here zero matrices) Interesting is also fact that probably every matrix (of even dimension) can be decomposed into symetrical part which has only real eigenvalues and skew-symmetrical which has only pure imaginary values what makes interesting analogy to complex numbers and their two parts, but I don't know whether there are importatnt consequences of this fact.

glS
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Widawensen
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2 Answers2

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It's because $-A^2 = A^T A$ has only real nonnegative eigenvalues: if $v$ is an eigenvector of $A$ with eigenvalue $\lambda$ then $$-\lambda^2 \|v\|^2 = -v^T A^2 v = v^T A^T A v = \|A v \|^2.$$

Skew-symmetric matrices do not have to have nonzero determinant, the zero matrix is a counterexample.

user356288
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  • Let's exclude zero matrices from our problem. It is so special that it could be at the same time formally symmetric and skew-symmetric what gives really no information about its nature. – Widawensen Jul 26 '16 at 07:52
  • @Widawensen OK but there are plenty of other examples of skew-symmetric matrices with determinant $0$, including in even dimension – user356288 Jul 26 '16 at 07:54
  • @ user356288, hmm, do you point me some example ? Additionally, such even dimensional skew-symmetric matrices with zero determinant have some special property which other don't have? – Widawensen Jul 26 '16 at 08:10
  • The equation written by you, dear user356288, provides so much information about the problem, probably it gives also a hint why odd dimensional skew-symmetric matrices must have determinant equal 0. Only for one eigenvalue = 0 in this case we have fulfilled this equation. Maybe we should have two eigenvalues for even dimensions equal 0 in order to have non-invertible matrix ? Is it possible? – Widawensen Jul 26 '16 at 08:16
  • Ok. I've found counterexample. It has two 0 eigenvalues.

    $\begin{bmatrix} 0 & 1 & 0 & 0 \ -1 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \end{bmatrix}$

    It's seems that such non-invertible matrix must have some even dimensional sub-matrix on diagonal as the zero matrix.

     – Widawensen Jul 26 '16 at 08:30
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Say A is a skew symmetry matrix
Say $\lambda$ is a real eigenvalue of A, then there exists a real vector $v≠0$ such that $Av=\lambda v$

By skew symmetry we have:
$v^TAv=(v^TAv)^T$ (this holds because the $v^TAv$ is a real and reals are equal to their transpose)
$=v^TA^T(v^T)^T=v^TA^Tv=v^T(-A)v=-v^TAv$
Thus $v^TAv=-v^TAv=0$

Thus $0 = v^TAv=v^T\lambda v = \lambda v^Tv = \lambda |v|^2$
(this last equality holds because v is real)
Thus $0=\lambda |v|^2$
Thus $\lambda=0$ or $v=0$, however $v≠0$ thus $\lambda=0$

Thus the only possible real eigenvalue of A is $0$

Stated differently, all non zero eigenvalues of a skew symmetric matrix have non zero imaginary component

Mathew
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