# real matrix fails to have an eigenvalue

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 $\pm i$. – Henry T. Horton Dec 31 '12 at 17:10
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? – Val Dec 31 '12 at 17:13
You have to fix at first the field of coefficients. – Sigur Dec 31 '12 at 17:14
But who says that the matrix coefficients are the same type as the EV? – Val Dec 31 '12 at 17:15
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) – Henry T. Horton Dec 31 '12 at 17:17

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. – Val Dec 31 '12 at 17:10
@Val, I don't see any problem with that quotation. I saw that you'd edited your post. – Sigur Dec 31 '12 at 17:12
Then, you should be able to see a problem in my interpretation. – Val Dec 31 '12 at 17:14
@Val, I would say that you should be able to see a problem after our comments. – Sigur Dec 31 '12 at 17:19
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! – Emily Dec 31 '12 at 17:59

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. – Erick Wong Dec 31 '12 at 17:59
An eigenvector should be non-zero too. – Geoff Robinson Dec 31 '12 at 18:44