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Doing some linear algebra exercises I found that:

Given $f \in \mathcal{End}(V)$ we define $f^\star$ an endomorphism such that, given $\phi$ a positive scalar product: $\phi(f(x),y)=\phi(x,f^\star(y))$.

Let $V$ a finite dimensional vector space, $\dim V=n$. Let $\phi$ a positive scalar product. Let $f \in \mathcal{End}(V)$ such that $$\sum\limits_{\lambda \in Sp(f)} m_\lambda=n,$$ where $m_\lambda$ is the algebraic multiplicity of the eigenvalue $\lambda$.

Prove that $$f=f^\star\iff\operatorname{trace}(ff^{\star})=\sum\limits_{\lambda \in Sp(f)} m_{\lambda}|\lambda|^2.$$

I did the $(\Rightarrow)$ using the spectral theorem and the diagonal form of $f$, but I can't handle the $(\Leftarrow)$, any suggestions?

snulty
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JCF
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    This result is only valid if the eigenvalues of $f$ are real. For example, If $f$ is diagonal with complex eigenvalues then $f\neq f^$, but $trace(ff^{\star})=\sum_{\lambda \in Sp(f)} m_{\lambda}|\lambda|^2.$ Now, the right result is the following: $f$ is normal iff $trace(ff^{\star})=\sum_{\lambda \in Sp(f)} m_{\lambda}|\lambda|^2.$ Notice that if $f$ is normal with real eigenvalues then $f=f^$. You can find the proof of this result here: http://math.stackexchange.com/questions/1030016/show-that-a-in-mathbbc-n-is-normal-iff-traa-sum-i-1n-lambda/1030026#1030026 . – Daniel Aug 20 '16 at 09:28
  • Thank you. It was missed the hypothesis that the eigenvalues are real in the text of the exercise. – JCF Sep 09 '16 at 07:14

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