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Prove that all of the eigenvalues of skew-symmetric matrix are complex numbers with the real part equal to $0$. Has anyone got a clue how to do it?

glS
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Michał
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    The tag ([tag:theorem-provers]) is for questions about software designed for checking formal proofs or assisting with writing them, see the tag-wiki. It is not intended for all questions which are about proofs of theorems. – Martin Sleziak Feb 09 '15 at 14:41
  • Duplicate: https://math.stackexchange.com/q/57100/96384 – Torsten Schoeneberg Jan 21 '22 at 02:25
  • The link mentioned by Torsten is better in that it specifically states this is limited to real matrices. The accepted answer does not mention, even though must be assumed going from line 1 to line 2 - otherwise proof invalid. – kirk beatty Apr 26 '24 at 04:59

3 Answers3

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$S$ being skew-symmetric means $S+S^{*}=0$, therefore $\forall x$ holds $x^{*}(S+S^{*})x = 0$. Assume $v$ is an eigenvector, hence $Sv = \lambda v$ and $v\neq 0$. Then, $v^{*}(S+S^{*})v = v^{*}Sv + v^{*}S^{*}v = v^{*}\lambda v + (Sv)^{*} v = \lambda v^{*} v + \bar{\lambda} v^{*} v = (\lambda + \bar{\lambda})v^{*}v = 0$, which means $\mathrm{Re}(\lambda) =0$.

Dimitar Ho
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  • I see there is confusion regarding the definition here. Skew symmetric means $S^t=-S$. The claim $S+S^*=0$ made by you is only true when $S$ is skew-Hermitian. – Shivani Goel Apr 13 '25 at 10:39
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Hint: If $A$ is a skew-symmetric matrix then what can you say about the conjugate transpose of $B=iA$ ?

Bebop
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The eigenvalue problem for a skew-symmetric matrix $M$ is \begin{align}M\textbf{x} &=\lambda \textbf{x}\\ \left(M\textbf{x}^*\right)^T\textbf{x}&=\left(\lambda^*\textbf{x}^*\right)^T\textbf{x}\\ \left(\textbf{x}^*\right)^TM^T\textbf{x}&=\lambda^*\left(\textbf{x}^*\right)^T\textbf{x}\\ \left(\textbf{x}^*\right)^T\left(-M\right)\textbf{x}&=\lambda^*\left(\textbf{x}^*\right)^T\textbf{x}\\ \left(\textbf{x}^*\right)^T\left(-\lambda\textbf{x}\right)&=\lambda^*\left(\textbf{x}^*\right)^T\textbf{x}\\ -\lambda\|\textbf{x}\|^2&=\lambda^*\|\textbf{x}\|^2\\ -\lambda&=\lambda^*\\ \therefore \Re(\lambda)&=0 \end{align}

where we have used the notation $^*$ for complex conjugation and $^T$ for transposition. From line 3 to line 4, we use the property of a skew-symmetric matrix: $M^T=-M$. The conclusion is equivalent to saying that $\lambda$ is either $0$ or pure imaginary.

Demosthene
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  • I don't know where you gone wrong in proving above but the result is only true for real skew symmetric matrices,for counter example consider the matrix $\begin{pmatrix} 0 & i\ -i & 0\end{pmatrix}$, its eigen values 1,-1. – Ibrahim Jul 15 '20 at 13:06
  • @Drmosthene if matrix is not real than you cannot move from line 2 to line 3,if I am not wrong?let me know your views in this regard.... – Ibrahim Jul 15 '20 at 13:19
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    Hi @lbs indeed you have just given an example of a complex skew-symmetric matrix having real eigenvalues; the statement in OP's question is in fact only true for real skew-symmetric matrices (which I assume here in my answer). – Demosthene Jul 22 '20 at 13:28
  • you have not mentioned that you have taken real symmetric matrix,neither OP mentioned such things,so firstly we should have made OP's Question correct,that's why I raised this question – Ibrahim Jul 23 '20 at 14:07
  • Going from line 1 to line 2 does not work with complex matrix. Take matrix given by Ibrahim, eigenvalue 1 eigenvector ($ i,1)$ . So that is where it is assumed real by Demosthene. Agree with Ibrahim, why not just say so? – kirk beatty Apr 26 '24 at 04:51