533: For Finite Set of Points on Real Vectors Space, if for Point, Set of Subtractions of Point from Other Points Is Linearly Independent, It Is So for Each Point|T.B.P.

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description/proof of that for finite set of points on real vectors space, if for point, set of subtractions of point from other points are linearly independent, it is so for each point

Topics

About: vectors space

Starting Context

The reader knows a definition of linearly independent subset of module.

Target Context

The reader will have a description and a proof of the proposition that for any finite set of points on any real vectors space, if for one of the points, the set of the subtractions of the point from the other points is linearly independent, it is so for each point.

Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

Main Body

1: Structured Description

Here is the rules of Structured Description.

Entities:

V

\in {the real vectors spaces}

{p_{0}, . . ., p_{n}}

\subseteq V

//

Statements:

\exists p_{k} ({p_{0} - p_{k}, . . ., \hat{p_{k} - p_{k}}, p_{n} - p_{k}} is linearly independent)

, where the hat mark denotes that the component is missing

⟹

\forall p_{l} ({p_{0} - p_{l}, . . ., \hat{p_{l} - p_{l}}, p_{n} - p_{l}} is linearly independent)

.
//

2: Natural Language Description

For any real vectors space,

V

, and any finite set of points,

{p_{0}, . . ., p_{n}} \subseteq V

, if for a

p_{k}

{p_{0} - p_{k}, . . ., \hat{p_{j} - p_{k}}, p_{n} - p_{k}}

is linearly independent, where the hat mark denotes that the component is missing, for each

p_{l}

{p_{0} - p_{l}, . . ., \hat{p_{l} - p_{l}}, p_{n} - p_{l}}

is linearly independent.

3: Proof

Let us suppose that

{p_{0} - p_{k}, . . ., \hat{p_{k} - p_{k}}, p_{n} - p_{k}}

is linearly independent.

Let

l

be any

l \in {0, . . ., n}

and

l \neq k

and

J_{l} := {0, . . ., n} ∖ {l}

.

Let

\sum_{j \in J_{l}} (t^{j} (p_{j} - p_{l})) = 0

= \sum_{j \in J_{l}} (t^{j} (p_{j} - p_{k} + p_{k} - p_{l})) = \sum_{j \in J_{l}} (t^{j} (p_{j} - p_{k})) - \sum_{j \in J_{l}} (t^{j} (p_{l} - p_{k})) = \sum_{j \in J_{l}} (t^{j} (p_{j} - p_{k})) - (\sum_{j \in J_{l}} t^{j}) (p_{l} - p_{k})

. Then,

t^{j} = 0

for each

j \neq k

and

\sum_{j \in J_{l}} t^{j} = 0

, but

\sum_{j \in J_{l}} t^{j} = t^{k} = 0

. So, each

t^{j}

0

, which means that

{p_{0} - p_{l}, . . ., \hat{p_{l} - p_{l}}, p_{n} - p_{l}}