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What's the minimal polynomial of a normal endomorphism $\phi$ with eigenvalues $2, 2, 1+i, 1+i, 1-i, 1-i, 3$?

It is $\mu_\phi | (t-2)^2(t-1-i)^2(t-1+i)^2(t-3)$ but is there any more I can derive from the fact that $\phi \circ \phi^*=\phi^* \circ \phi$?

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    **Hint:** if $\phi$ is normal, then it's diagonalizable. What can you say about minimal polynomial of diagonalizable endomorphism?2012-07-28
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    So since $\phi$ is normal it's diagonalizable. The minimal polynomial of a diagonalizable endomorphism is product of disjoint linear factors, so with $\mu_\phi | (t-2)^2(t-1-i)^2(t-1+i)^2(t-3)$ it follows that $\mu_\phi= (t-2)(t-1-i)(t-1+i)(t-3)$. Right? But what if it's over a $\mathbb{R}$-field?2012-07-28
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    $\phi$ is an endomorphism of a finite dimensional inner product space?2012-07-28
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    Yes, I assume so.2012-07-28
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    @J.kolab: That's right! =p And we have $\mu_\Phi=(t-2)(t-3)(t-1-i)(t-1+i)=(t-2)(t-3)(t^2-2t+2)$, so this works over a $\mathbb{R}$-field.2012-07-28
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    So far so good. But I don't think this works over $\mathbb{R}$ since over $\mathbb{R}$ the minimal polynomial is not a product of distinct linear factors, right?2012-07-28
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    @J.kolab The fact that you have listed non-real eigenvalues already assumes that you are working over $\Bbb{C}$. It ***does not*** make sense to work over a real vector space and talk about non-real eigenvalues. If you are working over $\Bbb{R}$ then you would say that the only eigenvalues of your endomorphism are $2,2$ and $3$.2012-07-29
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    @J.kolab: In general the minimal polynomial is the same if you consider over $\mathbb{R}$ or $\mathbb{C}$ (or in general, over an field or over the algebraically closure). Indeed, let $p_R$ and $p_C$ be the minimal polynomial of $\Phi$ on a space over $\mathbb{R}$ and the same space considered as over $\mathbb{C}$, respectively. Then $p_C$ divides $p_R$ (why?). At other hand, if $p_C(x)=x^n+a_1x^{n-1}+\cdot+a_n$, then $\phi^n+a_1\phi^{n-1}+\cdots+a_nI=0$, and if this has solution, the solutions are reals (try to see) (the unknows are $a_1,\cdots,a_n$ and $\phi$ is a real matrix). So $p_C|p_R$.2012-07-29
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    @J.kolab: as BenjaLim commented, if $\phi$ is an endomorphism in a $\mathbb{C}$-vector space, not always make sense see $\phi$ over an $\mathbb{R}$-space. But the reverse makes sense (and I assumed this in general case), and in you case we can see $\phi$ in a $\mathbb{R}$-space.2012-07-29

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An endomorphism $\phi$ is diagonalizable iff its minimal polynomial is a product of distinct linear factors (see http://en.wikipedia.org/wiki/Diagonalizable_matrix). A normal endomorphism is always diagonalizable over $\mathbb{C}$.

The minimal polynomialof $\phi$ must thus bu $(t-2)(t-1-i)(t-1+i)(t-3)$ because, as you pointed out, it divides $(t-2)^2(t-1-i)^2(t-1+i)^2(t-3)$ and shares that same roots as the characteristic polynomial of $\phi$.

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    What is if we're working over $\mathbb{R}$? Then the minimal polynomial is not a product of distinct linear factors and so the matrix is not diagonalizable?2012-07-28
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    Why is it that every normal endomorphism is diagonalizable? I think we didn't proof this proposition..2012-07-29
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    @J.kolab to prove that a normal endomorphism $\phi$ of $\mathbb{C}^n$ is diagonalizable, first prove that if a $V$ is a subspace of $\mathbb{C}^n$ which is stable by $\phi$, then $V^{\bot}$ is also stable by $\phi$. Then prove the result by induction on $n$.2012-07-29
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    @J.kolab Over $\mathbb{R}$, the endomorphism $\phi$ is not diagonalizable (it has non-real eigenvalues). Its minimal polynomial is thus not a product of disjoint linear factors, and it divides $(t-2)(t-1-i)(t-1+i)(t-3)$. So it should be $(t-2)(t^2-2t+2)(t-3)$.2012-07-29