Wikipedia:Reference desk/Archives/Mathematics/2011 October 10

= October 10 =

Transformation of a Gamma function
If 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 $$(-1/2,\infty)$$, it has an inverse function (for this domain), meaning that $$f(y)=x$$ (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.
Integer 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 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
If $$A$$ is a non-diagonalizable 2 x 2 matrix, prove that there nonetheless exists a matrix $$P$$ such that $$P^{-1} A P = \begin{bmatrix} \lambda & 1 \\ 0 & \lambda \\ \end{bmatrix}$$

$$A$$ being non-diagonalizable means that it must have exactly 1 eigenvalue.

$$AP = P\begin{bmatrix} \lambda & 1 \\ 0 & \lambda \\ \end{bmatrix}$$

$$\iff [Ap_1, Ap_2] = [\lambda p_1, p_1+\lambda p_2]$$

This shows that the first column of P is an eigenvector of A if $$\lambda$$ 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 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 $$(A-\lambda I)^2=0$$. A is not diagonalizable so for some $$p_2$$ you have $$(A-\lambda I)p_2\neq0$$. Denote $$p_1=(A-\lambda I)p_2$$, then $$(A-\lambda I)p_1=(A-\lambda I)^2p_2=0$$. $$p_1,\ p_2$$ are nonzero and one is not a multiple of the other (otherwise $$(A-\lambda I)p_2=0$$), so $$P=[p_1\ p_2]$$ satisfies the requirements. -- Meni Rosenfeld (talk) 09:29, 10 October 2011 (UTC)

General form of a diagonalizable matrix.
For a 2 x 2 mattrix $$\begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}$$ what algebraic criteria in terms of $$a,b,c,d$$ are necessary and sufficient to ensure that there exists a real matrix $$P$$ such that $$P^{-1}\begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}P$$ is diagonal? What algebraic criteria in terms of $$a,b,c,d$$ are necessary and sufficient to ensure that there does not exist a real matrix $$P$$ such that $$P^{-1}\begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}P$$ is diagonal but there exists a complex matrix $$P$$ such that $$P^{-1}\begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}P$$ is diagonal?

I can get $$(a+d)^2 - 4(ad - bc) \ne 0$$ 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.

$$(a+d)^2 - 4(ad - bc)$$ 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 ($$a=d,\ b=c=0$$). This is easy to show - if $$P^{-1}AP=\lambda I$$ then $$A=P(\lambda I)P^{-1}=\lambda I$$. 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 $$(a+d)^2 - 4(ad - bc) \ne 0$$ OR $$a=d,\ b=c=0$$, then this is a necessary and sufficient condition to ensure that $$A$$ is diagonalizable. However it doesn't answer the question as to under what conditions $$A$$ is diagonalizable by a real, or nonreal, matrix. Although I suspect replacing $$(a+d)^2 - 4(ad - bc) \ne 0$$ with the stronger statement $$(a+d)^2 - 4(ad - bc) > 0$$ or $$(a+d)^2 - 4(ad - bc) < 0$$ respectively may achieve this, but I'm not sure. After all, is it possible for $$P$$ 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 $$P^{-1}AP$$ is a real diagonal matrix then there is also some real matrix $$Q$$ such that $$Q^{-1}AQ$$ 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 $$P^{-1}AP$$ is real and cannot be diagonal.
 * So: If $$\Delta>0$$, then A has two distinct real eigenvalues so it is diagonalizable over the reals; if $$\Delta=0$$ then it has a double eigenvalue so it is diagonalizable iff it is diagonal; if $$\Delta<0$$ 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)