description/proof of that for linear map between modules with same cardinality bases, map is bijection iff corresponding matrix is invertible
Topics
About: module
The table of contents of this article
Starting Context
- The reader knows a definition of linear map.
- The reader knows a definition of basis of module.
- The reader knows a definition of ring of \(n \times n\) ring matrices.
- The reader knows a definition of bijection.
- 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 any bijective linear map between any modules is a 'modules - linear morphisms' isomorphism.
- The reader admits the proposition that for any \(n \times n\) ring matrix, if and only if the matrix multiplied by each column or row is the original column or row, the matrix is the identity.
- The reader admits the proposition that any map is a bijection if and only if it has an inverse.
Target Context
- The reader will have a description and a proof of the proposition that for any linear map between any modules with any same cardinality bases, the map is a bijection if and only if the corresponding matrix with respect to the bases is invertible.
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:
\(R\): \(\in \{\text{ the rings }\}\)
\(M_1\): \(\in \{\text{ the } R \text{ modules } \}\), with any basis, \(B_1 = \{{b_1}^1, ..., {b_1}^d\}\)
\(M_2\): \(\in \{\text{ the } R \text{ modules } \}\), with any basis, \(B_2 = \{{b_2}^1, ..., {b_2}^d\}\)
\(f\): \(: M_1 \to M_2\), \(\in \{\text{ the linear maps }\}\)
\(M\): \(\in M_n (R)\), \(= \text{ the matrix }\) which corresponds to \(f\) with respect to \(B_1\) 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
//
Statements:
\(f \in \{\text{ the bijections }\}\)
\(\iff\)
\(M \in \{\text{ the invertible matrices }\}\)
//
2: Proof
Whole Strategy: Step 0: see what \(M\) is; Step 1: suppose that \(f\) is a bijection; Step 2: see that \(M\) is invertible; Step 3: suppose that \(M\) is invertible; Step 4: see that \(f\) is a bijection.
Step 0:
As \(f\) is linear, \(f\) is represented by \(M\), by 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.
\(M\) is determined by that \(f ({b_1}^j) = M^j_l {b_2}^l\).
Step 1:
Let us suppose that \(f\) is a bijection.
Step 2:
\(f\) is a 'modules - linear morphisms' isomorphism, by the proposition that any bijective linear map between any modules is a 'modules - linear morphisms' isomorphism.
So, there is the linear inverse, \(f^{- 1}: M_2 \to M_1\).
As \(f^{- 1}\) is linear, \(f^{- 1}\) is represented by an \(M' \in M_n (R)\), by 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.
As \(f^{- 1} \circ f = id_{M_1}\) is linear, \(f^{- 1} \circ f\) is represented by an \(M'' \in M_n (R)\), by 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.
\(f^{- 1} \circ f (m_1) = f^{- 1} \circ f ({m_1}_j {b_1}^j) = {m_1}_j M''^j_l {b_1}^l\), but it equals \(id_{M_1} (m_1) = m_1 = {m_1}_l {b_1}^l\), which means that \({m_1}_j M''^j_l = {m_1}_l\), which means that \({m_1}^t M'' = {m_1}^t\), so, \(M'' = I\), by the proposition that for any \(n \times n\) ring matrix, if and only if the matrix multiplied by each column or row is the original column or row, the matrix is the identity.
On the other hand, \(f^{- 1} \circ f (m_1) = f^{- 1} (f (m_1)) = f^{- 1} (f ({m_1}_j {b_1}^j)) = f^{- 1} ({m_1}_j M^j_l {b_2}^l) = {m_1}_j M^j_l M'^l_p {b_1}^p\), which means that \({m_1}_j M''^j_p = {m_1}_j M^j_l M'^l_p\), which implies that \(M''^j_p = M^j_l M'^l_p\) (take \({m_1}_j = 1\) and the other components \(0\)), which means that \(M'' = M M'\).
So, \(M M' = I\).
As \(f \circ f^{- 1} = id_{M_2}\) is linear, \(f \circ f^{- 1}\) is represented by an \(M'' \in M_n (R)\), by 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.
\(f \circ f^{- 1} (m_2) = f \circ f^{- 1} ({m_2}_j {b_2}^j) = {m_2}_j M''^j_l {b_2}^l\), but it equals \(id_{M_2} (m_2) = m_2 = {m_2}_l {b_2}^l\), which means that \({m_2}_j M''^j_l = {m_2}_l\), which means that \({m_2} M'' = {m_2}\), so, \(M'' = I\), by the proposition that for any \(n \times n\) ring matrix, if and only if the matrix multiplied by each column or row is the original column or row, the matrix is the identity.
On the other hand, \(f \circ f^{- 1} (m_2) = f (f^{- 1} (m_2)) = f (f^{- 1} ({m_2}_j {b_2}^j)) = f ({m_2}_j M'^j_l {b_1}^l) = {m_2}_j M'^j_l M^l_p {b_2}^p\), which means that \({m_2}_j M''^j_p = {m_2}_j M'^j_l M^l_p\), which implies that \(M''^j_p = M'^j_l M^l_p\) (take \({m_2}_j = 1\) and the other components \(0\)), which means that \(M'' = M' M\).
So, \(M' M = I\).
So, \(M\) is invertible with the inverse, \(M'\).
Step 3:
Let us suppose that \(M\) is invertible.
Step 4:
There is an \(M' \in M_n (R)\) such that \(M' M = I\) and \(M M' = I\).
\(f': M_2 \to M_1\) defined as \(f' ({m_2}_j {b_2}^j) = {m_2}_j M'^j_l {b_1}^l\) is linear, by 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.
\(f' \circ f (m_1) = f' (f (m_1)) = f' (f ({m_1}_j {b_1}^j)) = f' ({m_1}_j M^j_l {b_2}^l) = {m_1}_j M^j_l M'^l_p {b_1}^p = {m_1}_j (M M')^j_p {b_1}^p = {m_1}_j I^j_p {b_1}^p = {m_1}_j {b_1}^j = m_1\), which means that \(f' \circ f = id_{M_1}\).
\(f \circ f' (m_2) = f (f' (m_2)) = f (f' ({m_2}_j {b_2}^j)) = f ({m_2}_j M'^j_l {b_1}^l) = {m_2}_j M'^j_l M^l_p {b_2}^p = {m_2}_j (M' M)^j_p {b_2}^p = {m_2}_j I^j_p {b_2}^p = {m_2}_j {b_2}^j = m_2\), which means that \(f \circ f' = id_{M_2}\).
So, \(f'\) is an inverse of \(f\).
\(f\) is a bijection, by the proposition that any map is a bijection if and only if it has an inverse.
