Every set of k+1 elements is dependent.

[Guest2]

Let $S = \left\{x_1, \ldots, x_k\right\}$ be independent set consisting of $$k$$ elements in a linear space $$V$$ and let $l(S)$ be the subspace spanned by $S$. Then every set of $$k+1$$ elements in $$l(S)$$ is dependent.

I've gotten nowhere with this - but I'm posting my attempt nonetheless if only for the sake showing effort.

Since $$S$$ is independent there exist scalars $d_1, \ldots, d_k$ such that $$\sum_{1 \le i \le k}d_ix_i = 0 \implies d_1 = d_2 = \ldots = d_k ~~~~~~ (1)$$

Let $y_1, \ldots, y_{k+1} \in l(S)$, and let $c_1, \ldots, c_k$ be scalars not all zero. Then we can write:

$$\begin{align} & \begin{aligned} c_1y_1 & = \sum_{1 \le i \le k}a_{1, i} x_i, ~ c_2y_2 & = \sum_{1 \le i \le k}a_{2, i} x_i, ~ \ldots, ~ c_{k+1}y_{k+1} = \sum_{1 \le i \le k}a_{k+1, i}x_i \end{aligned} \end{align}$$

Where $a_{1, i}, \ldots, a_{k+1, i}$ are scalars not all zero. Therefore $$\displaystyle \sum_{1 \le i \le k+1}c_iy_i = \sum_{1 \le r \le k+1}\sum_{1 \le i \le k}a_{r,i}x_i ~~~~~~~~~~~ (2)$$

Now if $a_{r,i}$ are all zero, then $\displaystyle \sum_{1 \le i \le k}c_iy_i = 0$ and we're done.

If not, then it's too hard for me. I thought of letting letting $a_{1,i} = -d_1, ~ a_{2,i} = d_{2}, \ldots, a_{k, i} = (-1)^kd_k$ and finally $a_{k+1, i} = 0$ if $k$ is even, and $d_{k}$ if $k$ is odd. Then from $(1)$ and (2) we have $\displaystyle \sum_{1 \le i \le k+1}c_iy_i = \sum_{1 \le r \le k+1}\sum_{1 \le i \le k} (-1)^i d_i x_i$. But nothing cancels, as $x_1, \ldots, x_k$ are distinct.

 

[Deveno]

Re: Every set of k+1 elements is independent.

To use linear independence and linear dependence in proofs, it is crucial to have a firm understanding of what the definitions are, and what they mean.

The definition of linear dependence:

Given a set of vectors $\{v_1,\dots,v_n\}$ there exists a non-zero linear combination that equals the 0-vector.

That is, there exists $c_1,\dots,c_n$ NOT ALL 0, such that:

$c_1v_1 + \cdots + c_nv_n = 0$.

We have to make the caveat "not all 0" for the $c_i$, because if they were, it would trivially be true that:

$0v_1 + \cdots + 0v_n = 0$, and EVERY set of vectors would be linear dependent, rendering it useless as a concept.

What it means:

We don't need ALL the vectors $\{v_1,\dots,v_n\}$ to generate the linear span of linear combinations:

$\{a_1v_1 + \cdots + a_nv_n : a_1,\dots,a_n \in F\}$ (here $F$ is the underlying field of our vector space-it many applications this is $\Bbb R$ or $\Bbb C$ (but it doesn't HAVE to be).

In other words, we have some redundancy in our generating set. To see this, suppose we have a linear combination that sums to the 0-vector:

$c_1v_1 +\cdots + c_nv_n = 0$.

Suppose that we know $c_k \neq 0$ ($k$ might be 1, or $n$, but I will pretend it is between them, you can figure out the cases $k = 1$ and $k = n$ on your own) is the non-zero scalar coefficient.

Then:

$v_k = \dfrac{-c_1}{c_k}v_1 +\cdots \dfrac{-c_{k-1}}{c_k}v_{k-1} + \dfrac{-c_{k+1}}{c_k}v_{k+1} + \cdots + \dfrac{-c_n}{c_k}v_n$

(again, this formula actually presupposes $2 < k < n-1$, and again, you can puzzle out on your own these special cases).

So $v_k$ can be expressed as a linear combination of the remaining vectors, it is unnecessary to generate the span.

Now for linear independence.

The definition:

the set $\{v_1,\dots,v_n\}$ of vectors is linearly independent if the ONLY linear combination of these vectors is the 0-combination, that is:

$c_1v_1 + \cdots + c_nv_n = 0 \implies c_1 =\cdots = c_n = 0$ (these are "two different zeros", the first is the 0-vector, the second is the "field/scalar zero").

What it means: we need ALL of the $v_i$ to generate the linear span.

To see this, suppose that we could write:

$v_k = a_1v_1 + \cdots + a_{k-1}v_{k-1} + a_{k+1}v_{k+1} + \cdots + a_nv_n$ for some $k$.

(again, the same caveat about the "special cases of $k$" apply as above, the modifications needed are minor).

Then $a_1v_1 + \cdots + a_{k-1}v_{k-1} + (-1)v_k + a_{k+1}v_{k+1} + \cdots + a_nv_n = 0$

and since $-1 \neq 0$, this is a non-zero linear combination that sums to the 0-vector. But, from the definition, no such linear combination exists, so we have a contradiction. So our supposition that we could write our $v_k$ as a linear combination of the others must be false.

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Will post more later.

 

[Guest2]

Re: Every set of k+1 elements is independent.

Thank you, I'll study your post throughly.