# 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

## 2 Answers

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.

You ask in particular:

"If we have a set of linearly independent vectors then we can't get 0 vector after their linear combination, isn't it true?"

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.