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

Give a quadratic polynomial $p$ such that $p(E)=0$.

I know that the eigenvalue of $u$ is $v^*u$. And the eigenvalue of the orthognal complement of $u$ must be $0$. But I really have no idea how I can construct a polynomial here. I know that $p(\lambda)=\lambda^{n-1}(\lambda-v^*u)=0$. Can I know conclude that $p(E)=E^{n-1}(E-v^*u)=0$ ? And therefore $E^2-v^*uE=0$.

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Such that $p(E)=0$ for every $v$? Or for $u$ and $v$ both fixed? –  1015 Apr 23 '13 at 12:48
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1 Answer 1

Note that for fixed $v$ there are only the two eigenvalues, $v^* u$ and $0$, so $p(\lambda) = \lambda (\lambda - v^* u)$ is the (monic) quadratic s.t. $p(E(v)) = 0$. However this polynomial varies with $v$.

An explicit calculation may be overkill, but with $E = u v^*$:

$$ p(E) = u v^* (u v^* - (v^* u) I) = (v^* u) u v^* - (v^* u) u v^* = 0 $$

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