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 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 $B(x,y)=2\left( \sum_{i=1}^{n}x_{i}y_{i}\right)-\sum_{i=1}^{n-1}(x_{i}y_{i+1}+x_{i+1}y_{i})$ 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: 2,169 Thanks: 1141 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
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Quote:
 Originally Posted by romsek 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.
This isn't a problem. An inner product requires only that $B(x,x) \geq 0$ with equality only for $x= 0$.

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: 1,603 Thanks: 115 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. Thanks from bigchest01
 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: 1,603 Thanks: 115 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
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Quote:
 Originally Posted by zylo 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.
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
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Quote:
 Originally Posted by cjem 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.
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
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 Originally Posted by zylo 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?
This amounts to showing $\sum_{i=1}^{n} x_i^2 > \sum_{i=1}^{n-1}x_i x_{i+1}$ whenever $x = (x_1,\dots,x_n) \neq 0$ for all $n \geq 2$. This is clearly true for $n = 2$ (if either of $x_1$ or $x_2$ is $0$, then $x_1^2 + x_2^2 > 0 = x_1 x_2$; else $x_1 x_2 \leq \operatorname{max}(x_1^2,x_2^2) < x_1^2 + x_2^2$).

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

$$\sum_{i=1}^{n+1} x_i^2 - \sum_{i=1}^{n}x_i x_{i+1} = \underbrace{\left(\sum_{i=1}^{n} x_i^2 - \sum_{i=1}^{n-1}x_i x_{i+1}\right)}_{>0 ~ \text{by inductive hypothesis}} + \underbrace{\left(x_{n+1}^2 - x_n x_{n+1}\right)}_{\geq 0 ~ \text{as} ~ x_{n+1} ~ \geq ~ x_n} > 0,$$

which completes the induction.

 October 16th, 2018, 12:28 PM #10 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 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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