I can't understand a definition and a theorem about vector spaces and linear (in)dependency of a set of vectors

There is a definition in the book that says:

A set of n vectors {$$e_1, e_2, ..., e_n$$} is linearly independent if there do not exist real numbers $$a_1, a_2, ..., a_n$$, where at least one of the $$a_i$$ is not zero, such that $$a_1e_1+a_2e_2+...+a_ne_n=\vec0$$ Otherwise, the set {$$e_1, e_2, ..., e_n$$} is called linearly dependent

I can't get into it at all. If we have a set of linearly independent vectors then we can't get $$0$$ vector after their linear combination, isn't it true?

There's also a theorem that states

Given two nonzero vectors $$e_1$$ and $$e_2$$, if $$e_1 \cdot e_2 = 0$$, then $$e_1$$ and $$e_2$$ are linearly independent

but that seems like it says about a narrow case where both vectors are orthogonal but vectors $$v_1 = \begin{bmatrix}1, 0\end{bmatrix}$$ and $$v_2 = \begin{bmatrix}1, 1\end{bmatrix}$$ are linearly independent because their span is a 2D plane, however, their dot product is not equal to $$0$$

I'm not a strong mathematician, the most simplistic explanations are preferred

• Regarding your second question: The theorem says that orthogonal vectors are linearly independent, not that linearly independent vectors are orthogonal Nov 28 '21 at 8:25

The definition of linear independence that you gave is the classical one, but it doesn’t help much with intuition, in my view.

I think it’s better to start by defining linear dependence: a set of vectors is linearly dependent if one of them can be written as a linear combination of the other ones. That should make sense because it closely matches the meaning of “linearly dependent” in normal (non-mathematical) English. Then, of course a set of vectors is said to be linearly independent if they are not linearly dependent.

This definition is equivalent to the classical one you cited, but it makes a lot more sense (to me, anyway).

So, think of the three vectors $$a = (1,0,0)$$, $$b=(0,1,0)$$, and $$c=(2,5,0)$$ in $$\mathbb R^3$$. Obviously $$c = 2a + 5b$$. So these vectors are linearly dependent. It’s also true that $$2a + 5b - c = 0$$, so we have a (non-silly) linear combination that gives us zero, as required by the classical definition, but this is rather less intuitive, in my view.

In fact, in $$\mathbb R^3$$, three vectors are linearly dependent if and only if they are coplanar. This should make sense —- if $$c$$ lies in the same plane as $$a$$ and $$b$$, then $$c$$ can be written as a linear combination of $$a$$ and $$b$$. My example above is just a rather trivial example of three coplanar vectors. I expect you can invent more interesting ones.

If two vectors are orthogonal, then it’s pretty obvious that neither can be written as a multiple of the other, so they’re certainly linearly independent. But, as you point out, this is a very special case, and it’s easy to find examples of pairs of vectors that are linearly independent without being orthogonal.

This is not true, but it's not entirely wrong either. For simplicity consider a set of two vectors $$\{u,v\}$$. We certainly can write the $$0$$-vector, which I'll call "$$\overline{0}$$" to distinguish it from the $$0$$-scalar, as a linear combination of $$u$$ and $$v$$ even if they are linearly independent: just take $$0u+0v.$$
And of course this works for any set of vectors whatsoever. A set of vectors is linearly independent iff there is no non-silly way of producing $$\overline{0}$$ as a linear combination.
Separately re: the end of your question, you are right that that theorem is a narrow case, but $$(i)$$ sometimes narrow cases are useful and $$(ii)$$ the idea of orthogonality is separately quite important.