My Math Forum Inequality with max for independent vectors

 Linear Algebra Linear Algebra Math Forum

 April 19th, 2017, 09:22 AM #1 Member   Joined: Jun 2012 From: Uzbekistan Posts: 59 Thanks: 0 Inequality with max for independent vectors Hello everybody. I have a problem which i can't prove or give a counterexample. Let $\displaystyle x=\{x_n\}_{n=1}^{\infty}$ and $\displaystyle y=\{x_n\}_{n=1}^{\infty}$ be two linearly independent vectors in infinite dimensional linear space i.e. $\displaystyle \{x_n\}_{n=1}^{\infty}\neq \{\lambda y_n\}_{n=1}^{\infty}$ with following properties: 1) $\displaystyle \max\limits_{1\leq n} \{|x_n|\} = \max\limits_{1\leq n} \{|y_n|\} = \max\limits_{1\leq n} \{|a_{11}x_n+a_{12}y_n|\} = \max\limits_{1\leq n} \{|a_{21}x_n+a_{22}y_n|\}=1$ 2) $\displaystyle \max\limits_{1\leq n} \{|x_n+y_n|\} =\max\limits_{1\leq n} \{|(a_{11}+a_{21})x_n+(a_{12}+a_{22})y_n|\}$ 3)$\displaystyle \max\limits_{1\leq n} \{|x_n-y_n|\}=\max\limits_{1\leq n} \{|(a_{11}-a_{21})x_n+(a_{12}-a_{22})y_n|\}$. Where $\displaystyle a_{ij}$ real numbers such that $\displaystyle a_{11}a_{22}-a_{12}a_{21}\neq 0$. Is it true that |$\displaystyle a_{11}a_{22}-a_{12}a_{21}|\leq 1$? I could prove that the above is true in assumption $\displaystyle 1>a_{ij}>0$. For other cases i tried to find counterexample with help of Maple but since the conditions are many Maple couldn't solve a system of equations with max and absolute value. Can anybody help me with this problem? Thank's in advance.

 Tags independent, inequality, max, vectors

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post StillAlive Calculus 5 September 2nd, 2016 11:45 PM beesee Probability and Statistics 2 June 9th, 2015 01:52 PM rakmo Algebra 5 March 28th, 2013 06:48 AM sparse_matrix Linear Algebra 0 November 8th, 2012 12:40 AM varunnayudu Advanced Statistics 2 November 27th, 2010 08:08 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top