Wikipedia:Reference desk/Archives/Mathematics/2011 October 10
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October 10
editTransformation of a Gamma function
editIf I graph y=Gamma(x+1/2)/(sqrt(pi)*Gamma(x+1)), its a curve. But is there any transformation I can do to turn the graph into a graph of a straight line. For example, if I had y=10+ln(x), then I can plot exp(y) versus x and get a straight line. Is something similar possible for the above function? Thanks! - Looking for Wisdom and Insight! (talk) 01:17, 10 October 2011 (UTC)
- Certainly. Since Gamma(x+1/2)/(sqrt(pi)*Gamma(x+1)) is strictly decreasing on , it has an inverse function (for this domain), meaning that (clearly linear in x). I find it extremely unlikely that such a function can be determined explicitly though. Sławomir Biały (talk) 01:25, 10 October 2011 (UTC)
I would like to submit an article on the subject of Integer Factorization.
editInteger Factorization is currently a subject of WikiPedia. I have determined a factoring method that is significantly different than the other methods presented so far. I have prepared an article that includes a document in Microsoft Word and a spread sheet in MicroSoft Excel. I would like to know first, where in WikiPedia do I submit my article and second, How do I include a spread sheet? I initially started to do this in "My talk" but I could not get it to accept the spread sheet so I did not complete the work. Now I am not sure that "My talk" is the right place to do this. If you could direct me to the appropriate pages or instructions, I would appreciate it. Thank you in advance. Variddell (talk) 01:16, 10 October 2011 (UTC)
- It sounds as though you have some original research. Wikipedia is not the place to publish new results; Wikipedia is an encyclopedia that collects information that has already been published by reliable sources. Please see Wikipedia:No original research for more information. —Bkell (talk) 01:19, 10 October 2011 (UTC)
- A better place would be a math forum like http://mymathforum.com/. What Wikipedia calls "original research" is not permitted here. CRGreathouse (t | c) 20:22, 10 October 2011 (UTC)
Existence of near-diagonalization
editIf is a non-diagonalizable 2 x 2 matrix, prove that there nonetheless exists a matrix such that
being non-diagonalizable means that it must have exactly 1 eigenvalue.
This shows that the first column of P is an eigenvector of A if is the eigenvalue, which always exists. How do I show that the other column of P always exists? Widener (talk) 04:56, 10 October 2011 (UTC)
- Jordan_form#A_proof gives the proof for the general case. Or are you looking for something more elementary for this special case? -- Meni Rosenfeld (talk) 07:34, 10 October 2011 (UTC)
- I suppose I am. Widener (talk) 08:02, 10 October 2011 (UTC)
- You can do this using the Cayley–Hamilton theorem (which itself is straightforward to prove for the 2x2 case), giving you . A is not diagonalizable so for some you have . Denote , then . are nonzero and one is not a multiple of the other (otherwise ), so satisfies the requirements. -- Meni Rosenfeld (talk) 09:29, 10 October 2011 (UTC)
- I suppose I am. Widener (talk) 08:02, 10 October 2011 (UTC)
General form of a diagonalizable matrix.
editFor a 2 x 2 mattrix what algebraic criteria in terms of are necessary and sufficient to ensure that there exists a real matrix such that is diagonal? What algebraic criteria in terms of are necessary and sufficient to ensure that there does not exist a real matrix such that is diagonal but there exists a complex matrix such that is diagonal?
I can get ensures that the eigenvalues are distinct. This is a sufficient condition to ensure diagonalizability, but not a necessary one. I can not make any further progress.
is the discriminant of the characteristic polynomial.Widener (talk) 09:15, 10 October 2011 (UTC)
- It's fairly simple for the 2x2 case. If the eigenvalues are not distinct, then the matrix is diagonalizable iff it is already diagonal ( ). This is easy to show - if then . In other words, the whole notion of matrix similarity rests on the noncommutativity of multiplication, but scalar matrices behave like scalars and commute. -- Meni Rosenfeld (talk) 09:36, 10 October 2011 (UTC)
- OK thanks - so if a,b,c,d satisfy OR , then this is a necessary and sufficient condition to ensure that is diagonalizable. However it doesn't answer the question as to under what conditions is diagonalizable by a real, or nonreal, matrix. Although I suspect replacing with the stronger statement or respectively may achieve this, but I'm not sure. After all, is it possible for to be real even if the eigenvalues are nonreal, or vice versa? Widener (talk) 10:22, 10 October 2011 (UTC)
- If A is a real matrix, P is a complex matrix and is a real diagonal matrix then there is also some real matrix such that is diagonal (and also nonreal ones, for example any multiple of Q by a nonreal scalar). If A has any nonreal eigenvalue for real P is real and cannot be diagonal.
- So: If , then A has two distinct real eigenvalues so it is diagonalizable over the reals; if then it has a double eigenvalue so it is diagonalizable iff it is diagonal; if then it has two distinct complex conjugate eigenvalues, so it is diagonalizable with a complex P but not with a real P. -- Meni Rosenfeld (talk) 10:48, 10 October 2011 (UTC)
- OK thanks - so if a,b,c,d satisfy OR , then this is a necessary and sufficient condition to ensure that is diagonalizable. However it doesn't answer the question as to under what conditions is diagonalizable by a real, or nonreal, matrix. Although I suspect replacing with the stronger statement or respectively may achieve this, but I'm not sure. After all, is it possible for to be real even if the eigenvalues are nonreal, or vice versa? Widener (talk) 10:22, 10 October 2011 (UTC)