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Let $x \in \mathbb{C^n}\setminus \{ \vec{0} \}$ be fixed. Find all the eigenvalues and eigenvectors of the matrices $A_1=xx^*$ and $$A_2\ =\ \left(\begin{array}{cc} 0 & x^* \\ x & 0 \\ \end{array}\right)$$

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Are you sure, that $x\in \mathbb{C}$ and not something like $x\in\mathbb{C}^{N}$? –  Alex May 11 '13 at 22:31
    
Something like $\Bbb C^{N\times 1}$ or $\Bbb C^{1\times N}$ to be more precised. It makes all the difference for $A_1$. –  Git Gud May 11 '13 at 22:32
    
and maybe you could add a definition of your star. I suppose it's the transposed and complex conjugated of $A$...but –  Alex May 11 '13 at 22:35
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Hint: Note that $A_1v=xx^*v$ is always a multiple of $x$. –  Hagen von Eitzen May 11 '13 at 22:37
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up vote 1 down vote accepted

Simply stick to the definition of eigenvector. Let $w=\pmatrix{z\\ v}$ be an eigenvector of $A_2$ corresponding to an eigenvalue $\lambda$. Then $A_2w=\lambda w$ and hence \begin{align*} x^\ast v &= \lambda z,\tag{1}\\ zx &= \lambda v.\tag{2} \end{align*} There are two possibilities:

  1. $\lambda = 0$. It follows that $x^\ast v=z=0$ and $v\neq0,\lambda\neq0$. The equation $x^\ast v=0$ also shows that the eigenspace of $\lambda=0$ is $(n-1)$-dimensional.
  2. $\lambda\ne0$. Equation $(2)$ implies that $z\neq0$ and $v\neq0$ and $v$ is a scalar multiple of $x$. Choose $v = x$, then $\lambda=z$ and you can determine $\lambda$ from $(1)$ (there are two solutions).

Eigenvalues and eigenvectors of $A_1$ can be found in a similar manner.

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