1
$\begingroup$

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 $\phi \circ \phi^*=\phi^* \circ \phi$?

  • 0
    @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

1 Answers 1

1

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 $\phi$.

  • 0
    @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