Talk:Perron–Frobenius theorem
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- Please could you define (in this article or elsewhere) what you mean by a left and right eigenspaces and eigenvectors? Thanks. Lupin 04:21, 9 Mar 2005 (UTC)
- Hello there are some littel mistakes in the article i think. I corriged them but another user eraesed my changes:
- i just want to say that positive means >=0 and exsit at least one element not rqual to zero
- strictly positiv >0 means there is no elemnt of the vector or Matrix equal to 0
- non negative means >=0 that means for example the zero vektor too
these are just some mistakes
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- furthermore only the sum of the columns in a stochastic matrix is one and not the rows
etc etc etc.... user: مبتدئ
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- A recent edit said "positive" meant "= 0", so I reverted. I'm being told on my discussion page that that was an error. I see that there was one other small edit that I reverted along with it; I think it changed ">" to "≥" or something like that, and I'm not sure right now which is right. Michael Hardy 17:55, 26 September 2006 (UTC)
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- OK, now I see that it did not say "= 0"; it said ">= 0". If "greater than or equal to" was intended, then it should say "≥ 0". Michael Hardy 17:57, 26 September 2006 (UTC)
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The matrix (0 1 // 1 0) seems to be a counterexample to the statement of the Perron-Frobenius theorem for non-negative matrices as stated in this article. The eigenvalues of this matrix are 1 and -1, and |-1| is not strictly less than 1. -- Dave Benson, Aberdeen, 13 Feb 2007.
The theorem as stated applies only to positive matrices, that is, every entry is strictly > 0. Your matrix has zero entries, thus it won't apply. The theorem can be extended to any positive definite matrix, but your matrix is not positive definite (as its determinant is -1).Oops, missed that you were talking about the nonnegative section. Indeed you are right, something's funny there. -- dcclark (talk) 03:35, 13 February 2007 (UTC)
if one assumes that the entries of the matrix are only non-negative (which, for applications, is often the right hypothesis), then one needs extra assumptions for the theorem to be valid: irreducibility (to ensure that all entries of the eigenvector are (strictly) positive) and primitiveness (to avoid other eigenvalues with the same modulus).
