MHB Vector Spaces .... Linear Dependence and Indepence .... Basic Proof Required

[Math Amateur]

In Andrew McInerney's book: First Steps in Differential Geometry, Theorem 2.4.3 reads as follows:https://www.physicsforums.com/attachments/5252McInerney leaves the proofs for the Theorem to the reader ...

I am having trouble formulating a proof for Part (3) of the theorem ...

Can someone help ...

Peter

 

[Deveno]

Suppose $S$ has $n$ elements, so $S = \{v_1,v_2,\dots,v_n\}$.

If one of these, say, $v_n$ (we can always "re-organize" our set $S$, so that the vector that is a linear combination of the others is the last one), is a linear combination of the others, we have:

$v_n = c_1v_1 + c_2v_2 + \cdots + c_{n-1}v_{n-1}$, for some scalars (field elements) $c_1,\dots,c_{n-1}$.

Hence:

$c_1v_1 + c_2v_2 +\cdots + c_{n-1}v_{n-1} + (-1)v_n = 0$.

These scalars cannot all be $0$, since in any field $1 \neq 0$, hence $-1 \neq -0 = 0$.

So, by the *definition* of linear dependence, $S$ is a linearly dependent set.

One caveat: $n = 1$ doesn't work. Why? Because if $v_1 \neq 0$, the set $\{v_1\}$ is linearly independent, since if:

$c_1v_1 = 0$, from $v_1 \neq 0$, we must have $c_1 = 0$.

On the other hand, if $S$ is a linearly dependent set of $n$ vectors, that for some $c_1,\dots,c_n$ not ALL $0$, we have:

$c_1v_1 +\cdots + c_nv_n = 0$.

Choose any $c_j \neq 0$ (we have at least one).

Then $v_j = -\left(\dfrac{c_1}{c_j}\right)v_1 - \cdots - \left(\dfrac{c_{j-1}}{c_j}\right)v_{j-1} - \left(\dfrac{c_{j+1}}{c_j}\right)v_{j+1} - \cdots - \left(\dfrac{c_n}{c_j}\right)v_n$

which is a linear combination of the other $n-1$ vectors.

For example, the set $S \subseteq \Bbb R^3$ given by:

$S = \{(0,1,0),(0,1,0),(2,5,0)\}$ is linearly dependent.