# Why do skew-symmetric matrices always have purely imaginary eigenvalues?

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.

• Duplicate: math.stackexchange.com/q/57100/96384 Jan 21, 2022 at 2:24
• @TorstenSchoeneberg in the question above it is additional question about determinant which is not presented in the linked question Jan 21, 2022 at 10:23
• Well the assertion about the determinant in that additional question is wrong. Jan 21, 2022 at 15:43
• @TorstenSchoeneberg Ok. Det. of Sk-Sym even dimension matrix can be non-zero, for odd dimension is always zero. Jan 21, 2022 at 16:34
• Yes. Because all eigenvalues of any matrix with real entries are either real, or come in pairs of complex conjugates. Since in odd dimension they cannot all be pairs, at least one of them has to be real. The only number that is in $\mathbb R$ and in $i \mathbb R$ is $0$. Jan 21, 2022 at 18:43

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.

• 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. Jul 26, 2016 at 7:52
• @Widawensen OK but there are plenty of other examples of skew-symmetric matrices with determinant $0$, including in even dimension Jul 26, 2016 at 7:54
• @ 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? Jul 26, 2016 at 8:10
• 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? Jul 26, 2016 at 8:16
• 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. Jul 26, 2016 at 8:30

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