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Given a square matrix $A$, we denote the set of its eigenvalues with $\operatorname{Spec}(A)$. Is it true that $$ \operatorname{Spec}(A)=\operatorname{Spec}(A^T) $$ for every $A\in\mathcal M_n(\Bbb R)$ (I think nothing changes if we consider $\Bbb C$ instead of $\Bbb R$)? |
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$\det(\lambda I-A^T)=\det((\lambda I-A)^T)=\det(\lambda I-A)$. Conclusion ? |
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The trace of a matrix its invariant for similarity, and $A$ is similar to $A^T$ for all $A\in M(n, \mathbb R)$. Moreover $A\sim_{\mathbb R} B$ iff $A \sim_{\mathbb C} B $. |
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