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For a linear map $T:V\to V$ where $V$ is a finite dimensional inner product space over $\mathbb{C}$, I know the result $\chi_{T^*}=\overline{\chi_T}$ (where $T^*$ is the adjoint map for $T$). My problem is I can't find a proof for it... Do you know one?

Also, I'm guessing we can also prove $m_{T^*}=\overline{m_T}$, but again, can't find a proof anywhere or generate one myself...

Can you help me please? I really don't see...

(here $\chi$ is for characteristic polynomial and $m$ for the minimal one)

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2 Answers 2

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The minimal and characteristic polynomials of (a square matrix) $A$ are both invariant under transposition of$~A$. For the characteristic polynomial this is because the determinant is invariant, and for the minimal polynomial because clearly $P[A^T]=P[A]^T$ for every (complex) polynomial $P$.

So instead of $T^*$ you can work with its transpose $\overline T$. But $\chi_{\overline T}=\overline{\chi_T}$ and $m_{\overline T}=\overline{m_T}$ are obvious because complex conjugation is a ring automorphism of$~\Bbb C$.

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  • $\begingroup$ Are you talking about tranposition or conjugacy? I'm not familiar with ring automorphisms... Also, why can you work with the trasnpose instead? $\endgroup$
    – Noom
    Nov 28, 2014 at 14:08
  • $\begingroup$ When I say transposition, I mean transposition (reflection in the diagonal, nothing more). When I say complex conjugation I mean complex conjugation (of all entries, in case of a matrix). Not conjugacy, that's something quite different. Basically, I break down the adjoint (=transpose complex conjugate) operation into separate steps: transposition followed by complex conjugation. I give separate arguments why both steps leave the charpoly and minpoly unchanged. The proof for complex conjugation just uses $\overline{x+y}=\bar x+\bar y$ and $\overline{xy}=\bar x\,\bar y$ (ring morphism property). $\endgroup$
    – Marc van Leeuwen
    Nov 28, 2014 at 14:32
  • $\begingroup$ I think I see! I see why transposition does not alter the characteristic and minimal polynomials, now since the matrix of $T$ can be identified with the conjugate-transpose of the matrix for $T$, we look at transposition on $\overline{T}$, right? Thank you very much, I get it now! $\endgroup$
    – Noom
    Nov 28, 2014 at 14:33
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Hint: What is the relationship between the eigenvalues of $T$ and $T^*$?

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  • $\begingroup$ I'm guessing they're conjugates (guessing from knowing that self-adjoint maps only have real eigenvalues) but not sure I can prove it... If it's true there is still multiplicity to deal with isn't it? $\endgroup$
    – Noom
    Nov 28, 2014 at 14:15

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