$\begingroup$A set of orthonormal vectors is linearly independent.
Since I wouldn't memorize any such facts that is mentioned in my course book, what is the reason behind it and why are they linearly independent?
$\endgroup$44 Answers
$\begingroup$Assume $0 = \sum_{i=1}^n \alpha_i v_i$, where the $v_1, \dots, v_n$ are orthonormal.
Then (show this)
$$0 = \langle \sum_{i=1}^n \alpha_i v_i, v_j\rangle = \alpha_j.$$
This proves the claim.
$\endgroup$$\begingroup$A set of vectors is linearly independent if each of them is outside the space spanned by the others.
To make the explanation easier, let's just use a set of three vectors in $\mathbb{R}^3$. The extension to higher dimensions doesn't add much except a bunch of indices.
Three vectors $\mathbf{u}$, $\mathbf{v}$, $\mathbf{w}$ in $\mathbb{R}^3$ are linearly independent if no one of them lies in the plane spanned by the other two. So, $\mathbf{w}$ must not lie in the plane $P(\mathbf{u},\mathbf{v})$ spanned by $\mathbf{u}$ and $\mathbf{v}$, and so on.
But if the three vectors are orthonormal, then $\mathbf{w}$ is actually perpendicular to the plane $P(\mathbf{u},\mathbf{v})$, (because it’s perpendicular to both $\mathbf{u}$ and $\mathbf{v}$) so it's very far from lying in this plane.
So, we might say that orthornormal vectors are "extremely" linearly independent. In some sense, they are nowhere even close to being linearly dependent -- they are as independent as they could possibly be.
Vague fluffy arguments like this won't gain you many points in math tests, but might help you visualize and understand, so that you don't have to memorizing so many facts.
$\endgroup$$\begingroup$This is just a more complete answer (but basically the same as answered by @PhoemueX):
Let the vectors $v_1,\ldots,v_n$ be orthonormal. That is
$v_i \cdot v_j = \begin{cases}1 & i = j\\0 & i \neq j\end{cases}$.
You want to show that $\sum_{i=1}^n \alpha_i v_i = 0 \iff \alpha_i = 0 \forall i \in {1,\ldots,n}$.
Note that $\langle \sum_{i=1}^n \alpha_i v_i, v_j \rangle = \alpha_1 v_1 \cdot v_j + \ldots + \alpha_n v_n \cdot v_j = \alpha_j v_j \cdot v_j = \alpha_j$ for all $j \in {1,\ldots,n}$.
Case "$\Leftarrow$": Trivial.
Case "$\Rightarrow$": Suppose $\sum_{i=1}^n \alpha_i v_i = 0$. Then $0 = \langle \sum_{i=1}^n \alpha_i v_i, v_j \rangle = \alpha_j$ for all $j \in {1,\ldots,n}$.
$\endgroup$1$\begingroup$You have to think of it as each vector adding a new dimension in your space. For instance it is impossible to compose vectors in a plane (in a linear combination) and obtain a vector which goes out of that plane. That is the "geometrical" reason why these are independant: each new vector cannot be obtained by composing the other ones.
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