description/proof of that for finite-dimensional columns or rows vectors space and linearly independent subset, subset can be shrunk into number-of-elements-dimensional columns or rows vectors space by choosing components
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of %field name% vectors space.
- The reader admits the proposition that for any injective linear map between any modules, the image of any linearly independent subset of the domain is linearly independent.
- The reader admits Cramer's rule for any system of linear equations.
- The reader admits the proposition that the determinant of any square matrix over any field is nonzero if and only if the set of the columns or the rows is linearly independent.
Target Context
- The reader will have a description and a proof of the proposition that for any finite-dimensional columns or rows vectors space and any linearly independent subset, the subset can be shrunk into a number-of-elements-dimensional columns or rows vectors space by choosing some components.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(d'\): \(\in \mathbb{N}\)
\(V'_c\): \(= \{(r^1, ..., r^{d'})^t \vert r^j \in F\}\), \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(V'_r\): \(= \{(r^1, ..., r^{d'}) \vert r^j \in F\}\), \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(S'_c\): \(= \{({r_1}^1, ..., {r_1}^{d'})^t, ..., ({r_n}^1, ..., {r_n}^{d'})^t\}\), \(\in \{\text{ the linearly independent subsets of } V'_c\}\)
\(S'_r\): \(= \{({r_1}^1, ..., {r_1}^{d'}), ..., ({r_n}^1, ..., {r_n}^{d'})\}\), \(\in \{\text{ the linearly independent subsets of } V'_r\}\)
\(V_c\): \(= \{(r^1, ..., r^n)^t \vert r^j \in F\}\), \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(V_r\): \(= \{(r^1, ..., r^n) \vert r^j \in F\}\), \(\in \{\text{ the } F \text{ vectors spaces }\}\)
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Statements:
\(\exists \{j_1, ..., j_n\} \subseteq \{1, ..., d'\} (S_c := \{({r_1}^{j_1}, ..., {r_1}^{j_n})^t, ..., ({r_n}^{j_1}, ..., {r_n}^{j_n})^t\} \in \{\text{ the linearly independent subsets of } V_c\})\)
\(\land\)
\(\exists \{j_1, ..., j_n\} \subseteq \{1, ..., d'\} (S_r := \{({r_1}^{j_1}, ..., {r_1}^{j_n}), ..., ({r_n}^{j_1}, ..., {r_n}^{j_n})\} \in \{\text{ the linearly independent subsets of } V_r\})\)
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2: Note
It is not that any \(\{j_1, ..., j_n\} \subseteq \{1, ..., d'\}\) would do: for example, when \(F = \mathbb{R}\), \(d' = 3\), and \(S'_c = \{(0, 1, 0)^t, (0, 0, 1)^t\}\), \(\{1, 2\} \subseteq \{1, 2, 3\}\) does not do because \(\{(0, 1)^t, (0, 0)^t\}\) is not linearly independent, while \(\{2, 3\} \subseteq \{1, 2, 3\}\) does.
3: Proof
Whole Strategy: Step 1: take \(c_1 ({r_1}^1, ..., {r_1}^{d'})^t + ... + c_n ({r_n}^1, ..., {r_n}^{d'})^t = 0\) and see that that equals \(\begin{pmatrix} {r_1}^1 & ... & {r_n}^1 \\ ... \\ {r_1}^{d'} & ... & {r_n}^{d'} \end{pmatrix} \begin{pmatrix} c_1 \\ ... \\ c_n \end{pmatrix} = 0\); Step 2: apply Cramer's rule for any system of linear equations to conclude that the rank of the matrix is \(n\); Step 3: choose a linearly independent \(S_c\); Step 4: conclude for \(S_r\).
Step 1:
Let us take \(c_1 ({r_1}^1, ..., {r_1}^{d'})^t + ... + c_n ({r_n}^1, ..., {r_n}^{d'})^t = 0\).
That equals \(\begin{pmatrix} {r_1}^1 & ... & {r_n}^1 \\ ... \\ {r_1}^{d'} & ... & {r_n}^{d'} \end{pmatrix} \begin{pmatrix} c_1 \\ ... \\ c_n \end{pmatrix} = 0\), obviously.
\(S_c\)' s being linearly independent equals that equation's having only the \(\begin{pmatrix} c_1 \\ ... \\ c_n \end{pmatrix} = 0\) solution.
Step 2:
By Cramer's rule for any system of linear equations, the rank of the matrix is \(n\): if the rank was smaller than \(n\), at least 1 of \(c_j\) s can be taken arbitrarily, a contradiction.
Step 3:
So, the matrix can be rearranged such that the top-left \(n \times n\) submatrix is determinant nonzero.
Then, the \(n\) columns of the submatrix is linearly independent, by the proposition that the determinant of any square matrix over any field is nonzero if and only if the set of the columns or the rows is linearly independent.
The \(n\) columns is an \(S_c\).
Step 4:
As for \(S_r\), we can just take the transpose of \(S'_r\), \(S'_c \subseteq V'_c\), take an \(S_c\), and take the transpose of \(S_c\), \(S_r \subseteq V_r\).
