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Let $\textbf{A}^\ast = -\textbf{A}$

Prove $\textbf{A}$ has only complex eigenvalues

Proof: Let $\lambda \in \mathbb{C}\ and\ \textbf{A}\vec{v} = \lambda\vec{v}\quad \vec{v} \neq \vec{0}$

$$\textbf{A}\vec{v} = \lambda\vec{v}$$ $$\Rightarrow (\textbf{A}\vec{v})^\ast = (\lambda\vec{v})^\ast$$ $$\Rightarrow (\vec{v}^\ast\textbf{A}^\ast)=(\lambda^\ast\vec{v}^\ast) $$ $$\because \vec{v} \neq \vec{0}$$ $$\therefore\vec{v}^\ast\textbf{A}^\ast\vec{v}=\lambda^\ast\vec{v}^\ast\vec{v}$$ $$\because \textbf{A}^\ast = -\textbf{A}$$ $$\therefore \vec{v}^\ast(-\textbf{A})\vec{v} = \lambda^\ast\vec{v}^\ast\vec{v}$$ $$\Rightarrow -\vec{v}^\ast\textbf{A}\vec{v} = \lambda^\ast\vec{v}^\ast\vec{v}$$ $$\because \textbf{A}\vec{v} = \lambda\vec{v}$$ $$\therefore -\vec{v}^\ast\lambda\vec{v}=\lambda^\ast\vec{v}^\ast\vec{v}$$ $$\Rightarrow -\lambda\vec{v}^\ast\vec{v}=\lambda^\ast\vec{v}^\ast\vec{v}$$ $$\Rightarrow \lambda\langle\vec{v}^\ast,\vec{v}^\ast \rangle=\lambda^\ast\langle\vec{v}^\ast,\vec{v}^\ast\rangle$$ $$\because \vec{v} \neq \vec{0}$$ $$\therefore \vec{v}^\ast \neq 0$$ $$\Rightarrow \langle\vec{v}^\ast,\vec{v}^\ast\rangle \gt 0\quad(\because \vec{v}^\ast\vec{v} = \langle\vec{v}^\ast, \vec{v}^\ast\rangle)$$ $$\Rightarrow -\lambda = \lambda^\ast$$ $$\Rightarrow \lambda \in\mathbb{R}$$

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    $\begingroup$ Side note: if $A$ is skew Hermitian, $iA$ is Hermitian. Assuming you have already shown that Hermitian matrices have real eigenvalues, this provides a shorter proof that $A$ has purely imaginary eigenvalues. $\endgroup$ – icurays1 Aug 9 '13 at 4:48
  • $\begingroup$ Here is the same problem. $\endgroup$ – Mhenni Benghorbal Aug 9 '13 at 19:29
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First of all, I presume you mean to say $\mathbf A$ has only imaginary eigenvalues; what would an eigenvalue of a complex matrix be if not complex?

Second of all, you made the same mistake with the definition of an eigenvector that you did in your previous proof. Namely: $\vec v$ is an eigenvector if $\mathbf{\vec v}$ is non-zero and there is some $\lambda\in\mathbb C$ such that $\mathbf A \vec v = \lambda v$.

That is, we do not deduce $\vec v$ is non-zero from the nature of the matrix; it is supposed under the definition of an eigenvalue. We do not suppose that $\lambda \neq 0$ because for some eigenvectors, even with skew-Hermitian matrices, $\lambda$ can be zero. So, at the top of your proof, write

"Let $\vec v\neq 0$ and $\mathbf A \vec v=\lambda v$".

Instead. The rest seems fine.

EDIT: Also, note that $\vec v^*\vec v$ is a matrix of one entry, and so you should write

$\therefore \vec{v}^\ast \vec{v} \neq 0$

rather than

$\therefore \vec{v}^\ast \vec{v} \neq \vec{0}$

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  • $\begingroup$ Note that there are different definitions of eigenvectors. For example I know one where Eigenvectors are vectors satisfying $Av=\lambda v$ and the Eigenvalues of $A$ are all $\lambda$ such that the Eigenspace is at least 1-dimensional $\endgroup$ – Dominic Michaelis Aug 9 '13 at 5:05
  • $\begingroup$ @DominicMichaelis thank you, I didn't realize that. $\endgroup$ – Omnomnomnom Aug 9 '13 at 5:34
  • $\begingroup$ thx to point out I'm missing the important definition of $\vec{v} \neq \vec{0}$ and $\lambda \in \mathbb{C}$ $\endgroup$ – bsdshell Aug 9 '13 at 6:48

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