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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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    @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 $\vec{v}$").2011-02-12

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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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    You could also say that the matrix $\left [\mathbf{u}_1 \cdots \mathbf{u}_n\right ]$ is full rank.2011-02-12