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Wikipedia says:

Having an eigenvalue is an accidental property of a real matrix (since it may fail to have an eigenvalue), but every complex matrix has an eigenvalue.

Yet, IMO, real matrices are subclass of complex ones. So, even without having any mathematical degree I see that this cannot be true.

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    I don't know what you mean by "unconscious job," but the eigenvalues of matrix have to be scalars from the underlying scalar field.$$\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$$ has no eigenvalues if you consider it as a real matrix, but if you consider it as a complex matrix, it has eigenvalues $\pm i$.2012-12-31
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    What does it mean "consider it as complex"? Do you mean that eighenvalue is the same as the matrix? That is, do you say that I cannot have a complex EV for a real matrix?2012-12-31
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    You have to fix at first the field of coefficients.2012-12-31
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    But who says that the matrix coefficients are the same type as the EV?2012-12-31
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    The Wikipedia article you linked says it: "Namely, let V be any vector space with some scalar field K, and let T be a linear transformation mapping V into V. We say that a vector x of V is an eigenvector of T if (and only if) there is **a scalar λ in K** such that T(x) = λx." (Emphasis added)2012-12-31
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    If you are working with real matrix (linear transformation between real vector spaces, for example) and the characteristic polynomial has no real roots then your matrix has no real eingevalues (or your transformation has no eigenvalues). I believe that Henry's example above is very clear.2012-12-31
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    Yes, I see that *a scalar λ in K* answers my question. How can I accept the comment as answer?2012-12-31
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    You can write an answer by yourself, if you want. Regards and happy new year.2012-12-31

2 Answers 2

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Answer in the comment: The Wikipedia article you linked says it: "Namely, let $V$ be any vector space with some scalar field $K$, and let $T$ be a linear transformation mapping $V$ into $V$. We say that a vector $x$ of $V$ is an eigenvector of $T$ if (and only if) there is a scalar $\lambda\in K$ such that $T(x) = \lambda x$."

($\lambda\in K$) must be in bold

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    Lest this requirement that $\lambda \in K$ seem unnatural, the point is that both $\mathbf x$ and $T(\mathbf x)$ are in $V$, so the only way the equality can possibly hold is if $\lambda$ belongs to the field of coefficients.2012-12-31
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    An eigenvector should be non-zero too.2012-12-31
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Eigenvalues are roots of a polynomial. Not every real polynomial has real roots. But every complex polynomial has roots.

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    This is absolutely correct answer and has absolutely nothing to do with the answer to my topic. I am asking to evaluate a concrete definition, not give your own.2012-12-31
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    @Val, I don't see any problem with that quotation. I saw that you'd edited your post.2012-12-31
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    Then, you should be able to see a problem in my interpretation.2012-12-31
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    @Val, I would say that you should be able to see a problem after our comments.2012-12-31
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    This is actually a very good answer that should illuminate the algebraic nature of matrices and eigenvalues, and cast away the thought that "eigenvalues are just a thing that matrices have." This has an interesting consequence for applications everywhere -- the Abel-Ruffini theorem means that there is no non-iterative general solution to computing the eigenvectors of a square matrix of dimension greater than 4!2012-12-31
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    It says that we can always have complex EV. This does not explain why we cannot have complex EV for real matrix at all (or the connection is visible only for matematicians).2012-12-31
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    Would it be better to say "Not every polynomial has real roots, but every polynomial has at least one complex root"? I only mention it becuase I'm not sure the distinction between a "real polynomial" and a "complex polynomial" is clear, and if those terms are meant to be related to the coefficients of the polynomials in question, then of course we should remember that a polynomial with all real coefficients (would that be a "real polynomial"?) can have non-real complex roots.2012-12-31
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    The question is what the polynomial coefficient have to do with the requirement for the roots to be in the same field?2012-12-31