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Prove that if $\mathbf{u}$ and $\mathbf{v}$ are nonzero orthogonal vectors in $\Bbb R^n$ they are linearly Independent.

I've struggled with this a bit, here is what I know so far:

Suppose $\mathbf{u}$ and $\mathbf{v}$ are orthogonal. Then $\mathbf{u\cdot v}=0$ and $c_1\mathbf{u}+c_2\mathbf{v}=0$ is linearly Independent iff $c_1=c_2=0$

I know I need to end with $c_1=c_2=0$ but I can't find a path that reaches this conclusion. I feel like I need to use properties of the dot product to connect my first assumption to my second assumption but I'm lost on the way.

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What do you mean when you say "and $c_1u+c_2v=0$ is linearly Independent iff $c_1=c_2=0$"? One does not describe an equation as being "linearly independent" ; one describes a set of vectors as being linearly independent. (The adjective "linearly independent" can also be applied to an ordered list of vectors.) You want to show that the set $\{u,v\}$ is linearly independent. And this means that if $c_1$ and $c_2$ are scalars such that $c_1 u + c_2 v = 0$, then $c_1 = c_2 = 0$. – littleO Nov 5 '12 at 21:03
Certainly, thank you. What I was drawing from was the definition for a set of vectors being linearly independent. c_1 and c_2 are scalars such that c_1u + c_2v = 0 – Ben Anderson Nov 5 '12 at 21:11
up vote 9 down vote accepted

Then $c_1\mathbf{u}+c_2\mathbf{v}=0$ is linearly Independent iff $c_1=c_2=0$

This does not actually make sense: $c_1\mathbf{u}+c_2\mathbf{v}=0$ isn’t something that can be linearly independent. Linear independence is a property of sets of vectors; $c_1\mathbf{u}+c_2\mathbf{v}=0$ is an equation involving vectors, but it is not a set of vectors.

The correct statement is that the set $\{\mathbf{u},\mathbf{v}\}$ is linearly independent (or, more casually, that the vectors $\mathbf{u}$ and $\mathbf{v}$ are linearly independent) iff it has the following property:

if $c_1$ and $c_2$ are scalars such that $c_1\mathbf{u}+c_2\mathbf{v}=0$, then $c_1=c_2=0$.

Your problem, then, is to show that if $\mathbf{u}$ and $\mathbf{v}$ are orthogonal, and if $c_1$ and $c_2$ are scalars such that $c_1\mathbf{u}+c_2\mathbf{v}=0$, then $c_1=c_2=0$.

To do this, assume that $c_1$ and $c_2$ are scalars such that $c_1\mathbf{u}+c_2\mathbf{v}=0$. Now apply the hint that Ayman Hourieh gave in his answer to show that $c_1=0$ and $c_2=0$.

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Thank you, and also thank you for editing the initial question, this is the first day I've stumbled across the site and don't have any idea how to use the language. – Ben Anderson Nov 5 '12 at 22:22
@Ben: You’re welcome. You can find lots of information about the language here. – Brian M. Scott Nov 5 '12 at 22:23
@BrianM.Scott Could you please tell me why Ayman's hint is any useful? Why do we want to calculate this dot product? What does it tell us? All we know is that $u\cdot v=0$. I managed to prove it in a different way: Express $u$ as $u=-\frac{c_{2}}{c_{1}}v$. Substitute back to $u\cdot v=0$. We see that $c_2$ has to be equal 0. Now we do the same for $v$ and it turns out both $c_1$ and $c_2$ are zero. How to prove the other direction? – user4205580 Mar 23 '15 at 22:16
Oh, I see. The other direction is just wrong :) OK, but how to interpret Ayman's hint? – user4205580 Mar 23 '15 at 22:33
@user4205580: Suppose that $c_1\mathbf{u}+c_2\mathbf{v}=0$. Then $$0=(c_1\mathbf{u}+c_2\mathbf{v})\cdot\mathbf{u}=c_1(\mathbf{u}\cdot\mathbf{u})+‌​c_2(\mathbf{v}\cdot\mathbf{u})=c_1(\mathbf{u}\cdot\mathbf{u})+0\;.$$ so $c_1=0$. – Brian M. Scott Mar 23 '15 at 22:41

Hint: Calculate: $$ (c_1 \mathbf{u} + c_2 \mathbf{v}) \cdot \mathbf{u} $$

What do you conclude about $c_1$? Do something similar for $c_2$.

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Thank you, this clears things up considerably. – Ben Anderson Nov 5 '12 at 21:48
@BenAnderson Happy to help! – Ayman Hourieh Nov 5 '12 at 21:52

Just use contradiction. If $u,v$ are linearly dependent and non-zero, then $u = cv$ (note: $c$ is a function of your $c_1,c_2$). Then $u\cdot v \neq 0 \rightarrow$ contradiction.

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