description/proof of that for module with \(2\) bases of same finite cardinality and module endomorphism, transition of endomorphism matrices w.r.t. change of bases is this
Topics
About: module
The table of contents of this article
Starting Context
- The reader knows a definition of %ring name% module.
- The reader knows a definition of %structure kind name% endomorphism.
- The reader admits 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.
- 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.
- The reader admits the proposition that if any square ring matrix has an inverse, the inverse is the unique inverse.
Target Context
- The reader will have a description and a proof 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.
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:
\(R\): \(\in \{\text{ the rings }\}\)
\(M\): \(\in \{\text{ the } R \text{ modules }\}\)
\(n\): \(\in \mathbb{N} \setminus \{0\}\)
\(B\): \(\in \{\text{ the bases for } M\} = \{b^j \vert j \in \{1, ..., n\}\}\)
\(B'\): \(\in \{\text{ the bases for } M\} = \{b'^l = N^l_j b^j \vert l \in \{1, ..., n\}\}\)
\(f\): \(: M \to M\), \(\in \{\text{ the ring endomorphisms }\}\)
\(O\): \(= \text{ the matrix of } f \text{ with respect to } B\) 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
\(O'\): \(= \text{ the matrix of } f \text{ with respect to } B'\) 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:
\(O' = N O N^{-1}\)
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2: Note
This holds not only for any vectors space \(M\) but also for any module \(M\) as far as such bases exist.
3: Proof
Whole Strategy: Step 1: see that \(N\) is uniquely determined; Step 2: take \(N' \in M_n (R)\) such that \(b^j = N'^j_l b'^l\), and see that \(N' = N^{- 1}\); Step 3: see that \(f (b'^l) = N^l_j O^j_m N'^m_p b'^p\).
Step 1:
\(N\) 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.
Step 2:
There is the unique \(N' \in M_n (R)\) such that \(b^j = N'^j_l b'^l\), 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 \(l \in \{1, ..., n\}\), \(b'^l = N^l_j b^j = N^l_j N'^j_m b'^m\), which implies that \(N^l_j N'^j_l = 1\) and \(N^l_j N'^j_m = 0\) for each \(m \neq l\), which implies that \(N N' = I\).
For each \(j \in \{1, ..., n\}\), \(b^j = N'^j_l b'^l = N'^j_l N^l_m b^m\), which implies that \(\) and \(N'^j_l N^l_j = 1\) and \(N'^j_l N^l_m = 0\) for each \(m \neq j\), which implies that \(N' N = I\).
So, \(N'\) is an inverse of \(N\) and is the inverse, \(N^{- 1}\), by the proposition that if any square ring matrix has an inverse, the inverse is the unique inverse.
Step 3:
For each \(l \in \{1, ..., n\}\), \(f (b'^l) = f (N^l_j b^j) = N^l_j f (b^j)\), because \(f\) is linear, \(= N^l_j O^j_m b^m = N^l_j O^j_m N'^m_p b'^p = O'^l_p b'^p\), which implies that \(N^l_j O^j_m N'^m_p = O'^l_p\), which implies that \(N O N' = O'\).
So, \(O' = N O N^{- 1}\).
