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"Given $\vec{u}_1,\ldots ,\vec{u}_n$ mutually orthogonal non-zero vectors, explain why for $\vec{v}=c_1\vec{u}_1+\ldots +c_n\vec{u}_n$ $c_k=\frac{\vec{v} \cdot \vec{u}_k}{\vec{u}_k \cdot \vec{u}_k}$"

This I explained by dotting both sides with $\vec{u}_k$ and simplifying everything. However, the question I now have is, how using this achieved result, can I show that $\vec{u}_1,\ldots ,\vec{u}_n$ are linearly independent? I was thinking of saying that in accordance to $c_k=\frac{\vec{v} \cdot \vec{u}_k}{\vec{u}_k \cdot \vec{u}_k}$, every coefficient can be of only one fixed value, so there is no room for changing one at the expense of another (as one could with coefficients of linearly dependent vectors), but I am not sure if this is right, and whether I am phrasing this correctly. Thanks for your help!

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Whey you say that they are mutually orthogonal, this implies that they are linearly independent because you cannot project one basis vector onto any of the others. That is if $\vec{v}$ is $\vec{u}_k$, then $\vec{v} \cdot \vec{u}_i = 0$ for any $i \neq k$. –  Tpofofn Feb 12 '11 at 21:30
    
Presumably you mean $\vec{v}=c_1\vec{u}_1+...+c_n\vec{u}_n$ (plus signs, not commas)? Your reasoning is basically correct, but there's no need for the "room for changing one at the expense of another" part -- if you've shown that each of the $c_i$ in the linear combination is uniquely determined by $\vec{v}$, you've shown that the $\vec{u}_i$ are linearly independent -- that's all there is to it. –  joriki Feb 12 '11 at 21:30
    
@Commodore64: uses5157's reasoning in the question was closer to the definition of linear independence, which is that the coefficients of linear combinations are uniquely determined, or equivalently that there is no linear combination resulting in the zero vector, or equivalently that none of the vectors can be expressed as a linear combination of the others. –  joriki Feb 12 '11 at 21:33
    
@joriki: Yes that's what I meant, sorry. The problem, however, asks to use part (a) to explain why they are independent. It seems obvious to me that they are, but I am not sure how to use the part (a) bit (which is $c_k=\frac{\vec{v} \cdot \vec{u}_k}{\vec{u}_k \cdot \vec{u}_k}$) to show the independence. So if I understand you correctly, since the formula from part a shows that coefficient is dependent only on $\vec{v}$ and $\vec{u}_k$, that is enough to demonstrate the independence? –  LinAlgStudent Feb 12 '11 at 21:33
    
@user5157: Yes, though I wouldn't phrase it quite that way. Linear independence is a property of the set of vectors $\vec{u}_k$, so it doesn't really make sense to say that this independence is demonstrated by showing that the coefficients are only dependent on $\vec{v}$ and $\vec{u}_k$ (since the $\vec{u}_k$ are fixed). Rather, you've shown that the $\vec{u}_k$ are linearly independent by showing that (given these $\vec{u}_k$) the coefficients are uniquely determined by $\vec{v}$ (and in that sense "depend only on $\vec{v}$"). –  joriki Feb 12 '11 at 21:47
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To show they are linearly independent, you want to show that any linear combination equal to zero is the trivial linear combination. So, suppose you have a linear combination $$\alpha_1\mathbf{u}_1 + \cdots + \alpha_n\mathbf{u}_n = \mathbf{0}.$$ Now, using the formula you got, taking $\mathbf{v}=\mathbf{0}$, will give you the value of each $\alpha_i$, namely: $$\alpha_i = \frac{\mathbf{0}\cdot\mathbf{u}_i}{\mathbf{u}_i\cdot\mathbf{u}_i}.$$ What does that tell you about $\alpha_1,\ldots,\alpha_n$?

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I just realized this, thanks a lot! It all makes sense now. –  LinAlgStudent Feb 12 '11 at 21:39
    
You could also say that the matrix $\left [\mathbf{u}_1 \cdots \mathbf{u}_n\right ]$ is full rank. –  Tpofofn Feb 12 '11 at 22:52
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