description/proof of that for linear map between modules with finite bases, transition of map representative matrices w.r.t. changes of bases is this
Topics
About: module
The table of contents of this article
Starting Context
- The reader knows a definition of basis of module.
- The reader knows a definition of linear map.
- The reader admits the proposition that for any module with any basis, the components set of any element with respect to the basis is unique.
Target Context
- The reader will have a description and a proof of the proposition that for any linear map between any modules with any finite bases, the transition of the map representative matrices with respect to changes of bases with the bases-cardinalities conserved is this.
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:
\(R\): \(\in \{\text{ the rings }\}\)
\(M_1\): \(\in \{\text{ the } R \text{ modules }\}\)
\(M_2\): \(\in \{\text{ the } R \text{ modules }\}\)
\(n_1\): \(\in \mathbb{N} \setminus \{0\}\)
\(n_2\): \(\in \mathbb{N} \setminus \{0\}\)
\(B_1\): \(\in \{\text{ the bases of } M_1\} = \{{b_1}^j \vert j \in \{1, ..., n_1\}\}\)
\(B_2\): \(\in \{\text{ the bases of } M_2\} = \{{b_2}^j \vert j \in \{1, ..., n_2\}\}\)
\(B'_1\): \(\in \{\text{ the bases of } M_1\} = \{{b'_1}^l = {N_1}^l_j {b_1}^j \vert l \in \{1, ..., n_1\}\}\)
\(B'_2\): \(\in \{\text{ the bases of } M_2\} = \{{b'_2}^l = {N_2}^l_j {b_2}^j \vert l \in \{1, ..., n_2\}\}\)
\(f\): \(: M_1 \to M_2\), \(\in \{\text{ the linear maps }\}\)
\(O\): \(= \text{ the matrix of } f \text{ with respect to } B_1 \text{ and } B_2\) defined in the proposition that for any map from any module with any \(d_1\)-elements basis into any module with any \(d_2\)-elements basis over any same ring, the map is linear if and only if the map is represented by the \(d_1 \times d_2\) ring matrix
\(O'\): \(= \text{ the matrix of } f \text{ with respect to } B'_1 \text{ and } B'_2\) defined in the proposition that for any map from any module with any \(d_1\)-elements basis into any module with any \(d_2\)-elements basis over any same ring, the map is linear if and only if the map is represented by the \(d_1 \times d_2\) ring matrix
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Statements:
\(O' = N_1 O {N_2}^{- 1}\)
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2: Note
This holds not only for any vectors spaces \(M_1\) and \(M_2\) but also for any modules \(M_1\) and \(M_2\) as far as such bases exist.
This is a generalization of the proposition that for any module with any \(2\) bases of any same finite cardinality and any module endomorphism, the transition of the endomorphism matrices with respect to the change of the bases is this.
We deal only with the case that \(B_1\) and \(B'_1\) have the same cardinality and that \(B_2\) and \(B'_2\) have the same cardinality, which is always with the case at least when \(R\) is a division ring, by the proposition that for any module over any division ring with any finite basis, there is no basis that has any larger cardinality.
3: Proof
Whole Strategy: Step 1: see that \(N_1\) and \(N_2\) are uniquely determined; Step 2: take \(N'_1 \in M_{n_1} (R)\) such that \({b_1}^j = {N'_1}^j_l {b'_1}^l\), and see that \(N'_1 = {N_1}^{- 1}\); Step 3: see that \(f ({b'_1}^l) = {N_1}^l_j O^j_m {N'_2}^m_p {b'_2}^p\).
Step 1:
\(N_1\) is uniquely determined, by the proposition that for any module with any basis, the components set of any element with respect to the basis is unique.
\(N_2\) is uniquely determined, likewise.
Step 2:
There is the unique \(N'_2 \in M_{n_2} (R)\) such that \(b_2^j = {N'_2}^j_l {b'_2}^l\), by the proposition that for any module with any basis, the components set of any element with respect to the basis is unique.
For each \(l \in \{1, ..., n_2\}\), \({b'_2}^l = {N_2}^l_j b_2^j = {N_2}^l_j {N'_2}^j_m {b'_2}^m\), which implies that \({N_2}^l_j {N'_2}^j_l = 1\) and \({N_2}^l_j {N'_2}^j_m = 0\) for each \(m \neq l\), by the proposition that for any module with any basis, the components set of any element with respect to the basis is unique, which implies that \(N_2 N'_2 = I\).
For each \(j \in \{1, ..., n_2\}\), \(b_2^j = {N'_2}^j_l {b'_2}^l = {N'_2}^j_l {N_2}^l_m b_2^m\), which implies that \({N'_2}^j_l {N_2}^l_j = 1\) and \({N'_2}^j_l {N_2}^l_m = 0\) for each \(m \neq j\), by the proposition that for any module with any basis, the components set of any element with respect to the basis is unique, which implies that \(N'_2 N_2 = I\).
So, \(N'_2\) is an inverse of \(N_2\) and is the inverse, \({N_2}^{- 1}\), by the proposition that if any square ring matrix has an inverse, the inverse is the unique inverse.
Step 3:
For each \(l \in \{1, ..., n_1\}\), \(f ({b'_1}^l) = f ({N_1}^l_j b_1^j) = {N_1}^l_j f (b_1^j)\), because \(f\) is linear, \(= {N_1}^l_j O^j_m b_2^m = {N_1}^l_j O^j_m {N'_2}^m_p {b'_2}^p = O'^l_p {b'_2}^p\), which implies that \({N_1}^l_j O^j_m {N'_2}^m_p = O'^l_p\), by the proposition that for any module with any basis, the components set of any element with respect to the basis is unique, which implies that \(N_1 O N'_2 = O'\).
So, \(O' = N_1 O {N_2}^{- 1}\).
