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

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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 $(-1/2,\mathrm{\infty })$, 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 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

If $A$ is a non-diagonalizable 2 x 2 matrix, prove that there nonetheless exists a matrix $P$ such that $P^{-1}AP=\left[\begin{array}{cc}\lambda & 1\\ 0& \lambda \end{array}\right]$

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

$AP=P\left[\begin{array}{cc}\lambda & 1\\ 0& \lambda \end{array}\right]$

$⟺[A{p}_1,A{p}_2]=[\lambda p_1,p_1+\lambda p_2]$

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#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 $(A-\lambda I{)}^2=0$. A is not diagonalizable so for some $p_2$ you have $(A-\lambda I)p_2\ne 0$. Denote $p_1=(A-\lambda I)p_2$, then $(A-\lambda I)p_1=(A-\lambda I{)}^2{p}_2=0$. $p_1,p_2$ 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 $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ 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 $P^{-1}\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]P$ 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 $P^{-1}\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]P$ is diagonal but there exists a complex matrix $P$ such that $P^{-1}\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]P$ 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 $\mathrm{\Delta }>0$, then A has two distinct real eigenvalues so it is diagonalizable over the reals; if $\mathrm{\Delta }=0$ then it has a double eigenvalue so it is diagonalizable iff it is diagonal; if $\mathrm{\Delta }<0$ 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)