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Let $0 \ne u \in \mathbb{C}^n$ fixed and consider for every $v \in \mathbb{C}^n$ the matrix $E(v)=uv^*$.

Find all eigenvalues of $E(v)$.

$E(v)=uv^*= \begin{pmatrix} u_1\\ \vdots \\ u_n \end{pmatrix} \begin{pmatrix} {v_1}^* & \cdots &{v_n} ^* \end{pmatrix} =\left(v_1^*\begin{pmatrix} u_1\\ \vdots \\ u_n \end{pmatrix}| ... | v_n^*\begin{pmatrix} u_1\\ \vdots \\ u_n \end{pmatrix} \right) $

Is this correct? I would conclude from this that $E(v)$ has rank 1. How could I find the eigenvalues ? Does this mean that there are $n-1$ eigenvectors with eigenvalue 0 ? What could be the last eigenvalue ? I'm stuck here, a hint would be appreciated !

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up vote 1 down vote accepted

Hint: $uv^\ast$ has rank $1$, so it has $n-1$ zero eigenvalues. Now you can see what the remaining eigenvalue is by considering $uv^\ast u$.

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$uv^*u=(v^*u)u$ as $v^*u \in \mathbb{C}$. Therefore the eigenvalues are $v^*u$ and $0$. Is this correct ? – Kasper Apr 23 '13 at 12:29
@Kasper Yes, that's correct. – user1551 Apr 23 '13 at 16:54

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