The wikipedia article is too bad in explaining. Is there any theorem that summarises which operators or matrices can be written in Jordan normal form?
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$\begingroup$ If the matrices are over the field of complex numbers, then every square matrix is similar to one in Jordan canonical form. If the matrices are over the field of real numbers, then every matrix is similar to one in rational canonical form. For a general field, I don't know. $\endgroup$– Disintegrating By PartsJan 29, 2014 at 11:28
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$\begingroup$ The Wikipedia page states conditions under which a matrix can be put into Jordan form: "If the vector space is over a field K, then a basis on which the matrix has the required form exists if and only if all eigenvalues of M lie in K, or equivalently if the characteristic polynomial of the operator splits into linear factors over K. This condition is always satisfied if K is the field of complex numbers." $\endgroup$– Disintegrating By PartsJan 29, 2014 at 11:45
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$\begingroup$ That's too many words. Short and simple and less rigorous is fine with me. $\endgroup$– Physics_mathsJan 29, 2014 at 12:10
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$\begingroup$ In short, if you want to be able to put a matrix into Jordan form, then you need to be able to completely factor the characteristic polynomial into linear factors. This can't always be done if you're stuck with real numbers because sometimes the roots of the characteristic polynomial are complex, and you can't completely factor polynomials such as $x^{2}+1$ just using real numbers; that's why there's a "rational canonical form" for this case. Based on your remarks, I doubt that you care about the case of a general field $K$, but the same is true there, too. $\endgroup$– Disintegrating By PartsJan 29, 2014 at 15:15
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$\begingroup$ I'm into physics hence the sloppyness. Thanks for your comment. $\endgroup$– Physics_mathsJan 29, 2014 at 15:41
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1 Answer
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Every square matrix can be written in Jordan Canonical form. This asserts that every linear operator has a basis of generalized eigenvectors.
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$\begingroup$ Is it the same with Schur decomposition? $\endgroup$ Jan 29, 2014 at 9:45
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$\begingroup$ Yes, every square matrix is similar to an upper triangular matrix. $\endgroup$ Jan 29, 2014 at 11:41
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