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I was given the following question:

Let $V$ be an inner product space and let $u,v\in V$ be two nonzero vectors. Prove or disprove:

  • If $\langle u,v\rangle=0$, then $u,v$ are linearly independent.
  • If $u,v$ are independent, then $\langle u,v\rangle=0$.
  1. I know that $u,v$ are arthogonal if $\langle u,v\rangle = 0$. So, since $\langle u,v\rangle = 0$, and $u,v$ are non zero vectors can I claim linear independence between the vectors directly? And if so, how do I explain it?

  2. This just seems wrong... I don't see how linear independence leads to this vectors having inner product of zero, meaning they are orthogonal. Any help or direction would be very helpful.

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  • $\begingroup$ Look at some examples of inner-product spaces. E.g $R^2$ where the inner product of vectors $<a,b>$ and $<c,d>$ is $a c+b d.$ And review the def'n of inner product : The inner product $(u,u)$ is not $0$ if $u\ne 0.$ $\endgroup$ – DanielWainfleet Jun 12 '16 at 23:17
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  1. It is true that $\textbf{u}$ and $\textbf{v}$ are linearly independent with these assumptions, however, it is not sufficient to claim it based on intuition. You can show it as follows:

Suppose $\alpha\textbf{u} + \beta\textbf{v} = \textbf{0}$. Then

$$ 0 = \langle \textbf{v}, \textbf{0}\rangle = \langle \textbf{v}, \alpha\textbf{u} + \beta\textbf{v} \rangle = \alpha \langle \textbf{v}, \textbf{u}\rangle + \beta\langle\textbf{v}, \textbf{v}\rangle = 0 + \beta|\textbf{v}|^2. $$

You can conclude from here that $\beta = 0$ (why?). A similar calculation shows that $\alpha = 0$, from which you can conclude that $\textbf{u}$ and $\textbf{v}$ are linearly independent

  1. For part two it should be easy to come up with a counter example in $\mathbb{R}^2$ to find two linearly independent vectors that are not orthogonal. This will show the statement is false.
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  • $\begingroup$ Thanks allot, would it be enough to show that: c1*v + c2*u = 0, and since u,v are not zero vector, c1 = c2 = 0? which is the more simplistic definition of linear independence? $\endgroup$ – Michael Segal Jun 12 '16 at 23:25
  • $\begingroup$ I'm not sure what you mean by the more simplistic definition of linear independence. You seem to be using the one and only definition, as I used in my answer. You have to do something to show that $c_1, c_2 = 0$, as there are plenty of nonzero vectors that have a nontrivial linear combination equalling the zero vector, which means they are linearly dependent. $\endgroup$ – user333870 Jun 12 '16 at 23:28
  • $\begingroup$ Ok ok ok, got you.. I got confused, i can't assume c1 and c2 to be zero, their sum could be zero irrelevent. thanks! $\endgroup$ – Michael Segal Jun 12 '16 at 23:33
  • $\begingroup$ You're welcome. $\endgroup$ – user333870 Jun 12 '16 at 23:34
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No, you cannot claim linear independent directly. You have to prove it. if $\alpha u +\beta v=0$, then $$ 0=\langle \alpha u+\beta v,u\rangle=\alpha\,\langle u,u\rangle. $$ As $\langle u,u\rangle\ne0$, it follows that $\alpha=0$. A similar argument shows that $\beta=0$, and so $u,v$ are linearly independent.

The second statement is wrong indeed. Let $V=\mathbb R^2$ with the usual inner product, let $u=(1,0)$ and $v=(1,1)$. Then $u,v$ are linearly independent but $\langle u,v\rangle=1$.

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  • $\begingroup$ I wouldn't say "a similar argument shows $\beta$ is zero", the fact that $v\neq 0$ shows that $\beta =0$, since $\alpha u +\beta v = 0$ and we know that $\alpha=0$. $\endgroup$ – snulty Jun 12 '16 at 23:33
  • $\begingroup$ Thank you for your help, about your example for the second statement. why V=C^2? it would hold for R as well no? $\endgroup$ – Michael Segal Jun 12 '16 at 23:35
  • $\begingroup$ Yes, it's a matter of taste. I'll change it to $\mathbb R $. $\endgroup$ – Martin Argerami Jun 12 '16 at 23:38

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