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In a linear algebra book, I've found a corollary that says: If $A\in\Bbb C^{n\times n}$ and $λ_1,...,λ_n$ eigenvalues of $A$ (not necessarily distinct), then $λ_1λ_2...λ_n=detA$, and $λ_1+...+λ_n=TrA$

Does this corollary also hold for $A\in \Bbb R^{n\times n}$ if and only if $A$ has exactly $n$ eigenvalues (not necessarily distinct)? What conditions must apply for $A$ for this corollary to hold?

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    It is always true, but the $\lambda_i$ may not be always real. – 2017-02-21
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    Okay, so $detA$ and $TrA$ will still be equal to that, but I have to use the complex eigenvalues? – 2017-02-21
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    Yes. Some real matrices do not have *any* real eigenvalue. – 2017-02-21

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Yes, the theorem is true also for a matrix with real entries.

Note that if a real matrix has a complex eigenvalue $\lambda$ than also the conjugate $\bar \lambda$ is an eigenvalue (because the eigenvalues are the roots of the characteristic polynomial), so we are sure that the determinant and the trace are real numbers ( because $\lambda \bar \lambda$ and $\lambda +\bar \lambda$ are real numbers).