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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?

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    $\begingroup$ The 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. $\endgroup$ – Martin Sleziak Feb 9 '15 at 14:41
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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.

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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 $\mathrm{Re}(\lambda) =0$.

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Hint: If $A$ is a skew-symmetric matrix then what can you say about the conjugate transpose of $B=iA$ ?

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