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 October 14th, 2018, 06:47 AM #1 Senior Member   Joined: Jan 2016 From: Blackpool Posts: 104 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: 2,631 Thanks: 1470 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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 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 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 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 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 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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 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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 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 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 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 Tags product, show Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post srecko Linear Algebra 1 October 27th, 2016 12:25 PM X-Toxic Venomm New Users 1 December 27th, 2013 10:13 AM Schrodinger Complex Analysis 1 November 10th, 2011 02:55 PM notnaeem Real Analysis 4 August 16th, 2010 01:32 PM

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