2024-06-03

606: For 2 Decompositions of Vector with Common Constituent, Coefficients of Common Constituent Are Same if Common Constituent Is Not on Vectors Space Spanned by Other Constituents

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description/proof of that for 2 decompositions of vector with common constituent, coefficients of common constituent are same if common constituent is not on vectors space spanned by other constituents

Topics


About: vectors space

The table of contents of this article


Starting Context



Target Context


  • The reader will have a description and a proof of the proposition that for any 2 decompositions of any vector with any common constituent, the coefficients of the common constituent are same if the common constituent is not on the vectors space spanned by the other constituents.

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:
F: { the fields }
V: { the F vectors spaces }
v1: V
{v2,...,vk,v2,...,vl}: V
v: =t1v1+t2v2+...+tkvk=t1v1+t2v2+...+tlvlV
V: = the vectors space spanned by {v2,...,vk,v2,...,vl}, V
//

Statements:
v1V

t1=t1
//


2: Natural Language Description


For any field, F, any F vectors space, V, any v1V, any {v2,...,vk,v2,...,vl}V, any v=t1v1+t2v2+...+tkvk=t1v1+t2v2+...+tlvlV, and the vectors space spanned by {v2,...,vk,v2,...,vl}, VV, if v1V, t1=t1.


3: Note


{v2,...,vk} or {v2,...,vl} does not need to be linearly independent.

When V=R2, v1=(0,1), v2=(1,0), and v2=(1,1), v=(1,1)=1v1+1v2=2v1+v2, so, t1t1: the coefficient of v1 depends on the choice of the 2nd constituent, v2 or v2.

When V=R3, v1=(0,0,1), v2=(1,0,0), v3=(0,1,0), v2=(a,b,0), and v3=(c,d,0) (suppose that adbc0), v=(1,1,1)=1v1+1v2+1v3=1v1+1/(adbc)(dc)v2+1/(adbc)(b+a)v3, so, necessarily t1=t1.

What is the difference? It that because v1 is perpendicular to the xy plane on which v2,v3,v2,v3 are? No (I mean, that is a too harsh requirement): take v1=(1,0,1), which is not perpendicular to the xy plane, but still, necessarily t1=t1, because the z component is contributed only by v1 and t1=t1 is required in order for the z component to be right.

In the 1st place, the concept of 'perpendicular' requires an inner product, but why do we need an inner product?

In fact, the difference is that for the 1st case, v1 is on the vectors space spanned by v2,v2, while for the 2nd case (including when v1 is not perpendicular to the xy plane), v1 is not on the vectors space spanned by v2,v3,v2,v3.


4: Proof


Let us suppose that v1V.

t1v1+t2v2+...+tkvk=t1v1+t2v2+...+tlvl implies that t1v1t1v1=(t1t1)v1=t2v2...tkvk+t2v2+...+tlvl. As v1 is not on V, t1t1=0: otherwise, v1=1/(t1t1)(t2v2...tkvk+t2v2+...+tlvl), a contradiction.


References


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