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
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Note
- 4: Proof
Starting Context
- The reader knows a definition of %field name% vectors space.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(v_1\): \(\in V\)
\(\{v_2, ..., v_k, v'_2, ..., v'_l\}\): \(\subseteq V\)
\(v\): \(= t^1 v_1 + t^2 v_2 + ... + t^k v_k = t'^1 v_1 + t'^2 v'_2 + ... + t'^l v'_l \in V\)
\(V'\): \(= \text{ the vectors space spanned by } \{v_2, ..., v_k, v'_2, ..., v'_l\}\), \(\subseteq V\)
//
Statements:
\(v_1 \notin V'\)
\(\implies\)
\(t^1 = t'^1\)
//
2: Natural Language Description
For any field, \(F\), any \(F\) vectors space, \(V\), any \(v_1 \in V\), any \(\{v_2, ..., v_k, v'_2, ..., v'_l\} \subseteq V\), any \(v = t^1 v_1 + t^2 v_2 + ... + t^k v_k = t'^1 v_1 + t'^2 v'_2 + ... + t'^l v'_l \in V\), and the vectors space spanned by \(\{v_2, ..., v_k, v'_2, ..., v'_l\}\), \(V' \subseteq V\), if \(v_1 \notin V'\), \(t^1 = t'^1\).
3: Note
\(\{v_2, ..., v_k\}\) or \(\{v'_2, ..., v'_l\}\) does not need to be linearly independent.
When \(V = \mathbb{R}^2\), \(v_1 = (0, 1)\), \(v_2 = (1, 0)\), and \(v'_2 = (1, - 1)\), \(v = (1, 1) = 1 v_1 + 1 v_2 = 2 v_1 + v'_2\), so, \(t^1 \neq t'^1\): the coefficient of \(v_1\) depends on the choice of the 2nd constituent, \(v_2\) or \(v'_2\).
When \(V = \mathbb{R}^3\), \(v_1 = (0, 0, 1)\), \(v_2 = (1, 0, 0)\), \(v_3 = (0, 1, 0)\), \(v'_2 = (a, b, 0)\), and \(v'_3 = (c, d, 0)\) (suppose that \(a d - b c \neq 0\)), \(v = (1, 1, 1) = 1 v_1 + 1 v_2 + 1 v_3 = 1 v_1 + 1 / (a d - b c) (d - c) v'_2 + 1 / (a d - b c) (-b + a) v'_3\), so, necessarily \(t^1 = t'^1\).
What is the difference? It that because \(v_1\) is perpendicular to the \(x - y\) plane on which \(v_2, v_3, v'_2, v'_3\) are? No (I mean, that is a too harsh requirement): take \(v_1 = (1, 0, 1)\), which is not perpendicular to the \(x - y\) plane, but still, necessarily \(t^1 = t'^1\), because the \(z\) component is contributed only by \(v_1\) and \(t^1 = t'^1\) 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, \(v_1\) is on the vectors space spanned by \(v_2, v'_2\), while for the 2nd case (including when \(v_1\) is not perpendicular to the \(x - y\) plane), \(v_1\) is not on the vectors space spanned by \(v_2, v_3, v'_2, v'_3\).
4: Proof
Let us suppose that \(v_1 \notin V'\).
\(t^1 v_1 + t^2 v_2 + ... + t^k v_k = t'^1 v_1 + t'^2 v'_2 + ... + t'^l v'_l\) implies that \(t^1 v_1 - t'^1 v_1 = (t^1 - t'^1) v_1 = - t^2 v_2 - ... - t^k v_k + t'^2 v'_2 + ... + t'^l v'_l\). As \(v_1\) is not on \(V'\), \(t^1 - t'^1 = 0\): otherwise, \(v_1 = 1 / (t^1 - t'^1) (- t^2 v_2 - ... - t^k v_k + t'^2 v'_2 + ... + t'^l v'_l)\), a contradiction.
