October 14th, 2018, 06:47 AM  #1 
Senior Member Joined: Jan 2016 From: Blackpool Posts: 100 Thanks: 2  Show this example is an inner product
\[B(x,y)=2\left( \sum_{i=1}^{n}x_{i}y_{i}\right)\sum_{i=1}^{n1}(x_{i}y_{i+1}+x_{i+1}y_{i})\] I need help showing that this is positive definite as i think I have already shown it is symmetric and linear in the first variable. Thanks guys! 
October 15th, 2018, 10:32 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,262 Thanks: 1198 
Are there restrictions on $x$ and $y$? Fooling around with this using random numbers to fill $x$ and $y$, I was able to produce an example where $B(x,y) < 0$. Actually, fooling with it a bit more, I see it's pretty common that $B(x,y)<0$. It looks to occur approximately 10% of the time. Last edited by skipjack; October 15th, 2018 at 05:08 PM. 
October 15th, 2018, 12:41 PM  #3  
Senior Member Joined: Sep 2016 From: USA Posts: 531 Thanks: 304 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
Last edited by skipjack; October 15th, 2018 at 05:09 PM.  
October 15th, 2018, 02:58 PM  #4 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,621 Thanks: 117 
From Definition of Inner Product B(x,y)=B(y,x) B(ax+by,z)=aB(x,z)+bB(y,z) B(x,x)>0 if x $\displaystyle \neq$ 0, and B(0,0)=0 Easiest with a summation convention if it can be set up that way. if I get a chance. 
October 16th, 2018, 06:46 AM  #5 
Newbie Joined: Oct 2018 From: United Kingdom Posts: 2 Thanks: 0 Math Focus: Multiplication 
Hm yes I'll have a look later going to the gym to work on that chest right now! 
October 16th, 2018, 07:26 AM  #6 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,621 Thanks: 117 
Using summation convention over repeated indices: $\displaystyle B(x,y)= 2x_{i}y_{i}x_{i}y_{i+1}y_{i}x_{i+1}$ $\displaystyle B(x,y)=B(y,x)$ $\displaystyle B(x,x)=2x_{i}x_{i}2x_{i}x_{i+1}$ For n=3, $\displaystyle B(x,x)= 2(x_{1}^{2}+x_{2}^{2} +x_{3}^{2})2(x_{1}x_{2}+x_{2}x_{3})$ which is not always greater than zero and B(x,y) is not an inner product. You can check second condition for IP as an exercise. 
October 16th, 2018, 08:13 AM  #7 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 286 Thanks: 88 Math Focus: Number Theory, Algebraic Geometry  Of course $B(x,x)$ is not always positive  it is zero when $x$ is the zero vector. But it is positive whenever $x$ is not the zero vector, which is all we need.

October 16th, 2018, 09:27 AM  #8 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,621 Thanks: 117  It didn't look like B(x,x) was always positive when I first did it. On closer inspection it seems to be true. Can you prove it?

October 16th, 2018, 10:16 AM  #9  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 286 Thanks: 88 Math Focus: Number Theory, Algebraic Geometry  Quote:
Suppose it is true for some $n \geq 2$, and let $x = (x_1,\dots x_{n+1}) \neq 0$. Up to reordering the $x_i$, we may assume $x_{n+1} = \operatorname{max}(x_1,\dots,x_{n+1}$). Now either $(x_1, \dots, x_n) = 0$ (in which case the result is clearly true: we would have $\operatorname{LHS} = x_{n+1}^2 > 0 = x_n x_{n+1} = \operatorname{RHS}$) or $$\sum_{i=1}^{n+1} x_i^2  \sum_{i=1}^{n}x_i x_{i+1} = \underbrace{\left(\sum_{i=1}^{n} x_i^2  \sum_{i=1}^{n1}x_i x_{i+1}\right)}_{>0 ~ \text{by inductive hypothesis}} + \underbrace{\left(x_{n+1}^2  x_n x_{n+1}\right)}_{\geq 0 ~ \text{as} ~ x_{n+1} ~ \geq ~ x_n} > 0,$$ which completes the induction.  
October 16th, 2018, 12:28 PM  #10 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,621 Thanks: 117 
Ref preceding post. It doesn't work. You can't change induction premise by reordering. For example, suppose I have reordered to put $\displaystyle x_{2}$ in last place: $\displaystyle B(x,x) = x_{1}^{2}+ x_{3}^{2}+ x_{2}^{2}x_{1}x_{3}x_{3}x_{2}$ Now put $\displaystyle x_{2}$ back where it was $\displaystyle B(x,x) = x_{1}^{2}+ x_{2}^{2}+ x_{3}^{2}x_{1}x_{2}x_{2}x_{3}$ They aren't the same. Last edited by zylo; October 16th, 2018 at 12:44 PM. Reason: clarify 

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