
Linear Algebra Linear Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
March 11th, 2017, 05:52 PM  #1 
Member Joined: Mar 2017 From: venezuela Posts: 34 Thanks: 3  Prove this New Determinant Property (New HighOrder numerical methods)
I suspect this property does not appear in the math literature, so please kindly prove it: Given the square matrix $A$ with $k=n+1$ rows and columns:% \[ A=% \begin{pmatrix} x_{1} & x^{n/k} & x^{(n1)/k} & x^{(n2)/k} & x^{(n3)/k} & \cdots & {x}^{3/k} & {x}^{2/k} & {x}^{1/k}\\ 0 & {x}_{2} & x^{n/k} & x^{(n1)/k} & x^{(n2)/k} & \cdots & {x}^{4/k} & {x}^{3/k} & x^{2/k}\\ 0 & 0 & x_{3} & x^{n/k} & x^{(n1)/k} & \cdots & {x}^{5/k} & {x}^{4/k} & {x}^{3/k}\\ 0 & 0 & 0 & {x}_{4} & x^{n/k} & \cdots & {x}^{6/k} & {x}^{5/k} & {x}^{4/k}\\ 0 & 0 & 0 & 0 & {x}_{5} & \cdots & {x}^{7/k} & {x}^{6/k} & {x}^{5/k}\\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\ 0 & 0 & 0 & 0 & 0 & \cdots & {x}_{n1} & x^{n/k} & {x}^{(n1)/k}\\ 0 & 0 & 0 & 0 & 0 & \cdots & 0 & {x}_{n} & {x}^{n/k}\\ x^{n/k} & x^{(n1)/k} & x^{(n2)/k} & x^{(n3)/k} & x^{(n4)/k} & \cdots & {x}^{2/k} & {x}^{1/k} & 1 \end{pmatrix} \] $\ $ \[ \left\vert A\right\vert =\ \ (1)^{n}\ \left( xx_{1}\right) \left( xx_{2}\right) \left( xx_{3}\right) \cdots\left( xx_{n}\right) \] Best regards, www.domingogomezmorin.wordpress.com . 
March 12th, 2017, 05:52 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,879 Thanks: 766 
That looks like a straight forward "expansion by minors". Have you tried proof by induction on n? If n= 1, so k= 2, then this is the determinant = (x x_1)[/tex] so the theorem is true for n= 1. Now try doing it for n= 2 so k= 3. Do you see how this determinant is related to the first matrix? 
March 14th, 2017, 08:38 AM  #3 
Member Joined: Mar 2017 From: venezuela Posts: 34 Thanks: 3 
Many thanks, indeed. I already did a proof, however, I thought this matrix could be of interest to some people and some other ideas could arise from this. This matrix is based on new highorder rootapproximating algorithms based on the Arithmonic Mean (Particular case of the Generalized Mediant). Domingo Gomez Morin 
March 14th, 2017, 09:14 AM  #4  
Senior Member Joined: Sep 2015 From: USA Posts: 1,660 Thanks: 844  Quote:
One post is more than enough.  

Tags 
determinant, highorder, methods, numerical, property, prove 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Highorder rootapproximating methods for children  arithmo  Number Theory  9  March 7th, 2017 12:09 PM 
Highorder root approximating methods for children  arithmetic  Number Theory  0  October 9th, 2016 08:19 PM 
New highorder rootapproximating methods for children  arithmetic  Number Theory  0  September 11th, 2016 03:43 PM 
Hello, why is the determinant turns out different using two different methods ?  d2idan  Linear Algebra  2  December 23rd, 2014 01:01 PM 
Unknown highorder arithmetical methods.  mirror  Applied Math  2  September 8th, 2013 09:04 AM 