description/proof of that for map from module with \(d_1\)-elements basis into module with \(d_2\)-elements basis over same ring, map is linear iff map is represented by \(d_1 \times d_2\) ring matrix
Topics
About: module
About: matrices space
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 knows a definition of %ring name% matrices space.
- 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 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.
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, \(B_1 = \{{b_1}^1, ..., {b_1}^{d_1}\}\)
\(M_2\): \(\in \{\text{ the } R \text{ modules }\}\), with any basis, \(B_2 = \{{b_2}^1, ..., {b_2}^{d_2}\}\)
\(f\): \(: M_1 \to M_2\)
\(M\): \(\in \{\text{ the } d_1 \times d_2 R \text{ matrices }\}\), such that \(f ({b_1}^j) = M^j_l {b_2}^l\)
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Statements:
\(f \in \{\text{ the linear maps }\}\)
\(\iff\)
\(\forall m_1 = {m_1}_j {b_1}^j \in M_1 (f (m_1) = {m_1}_j M^j_l {b_2}^l)\)
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2: Note
\(\forall m_1 = {m_1}_j {b_1}^j \in M_1 (f (m_1) = {m_1}_j M^j_l {b_2}^l)\) means nothing but "\(f\) is represented by \(M\)", because \(M\) determines \(f\), because the decomposition, \({m_1}_j {b_1}^j\), is unique, by the proposition that for any module with any basis, the components set of any element with respect to the basis is unique.
It means that \(f (m_1)_l = {m_1}_j M^j_l\) where \(f (m_1)_l\) is the \(l\) component of \(f (m_1)\), so, \(\begin{pmatrix} f (m_1)_1 & ... & f (m_1)_{d_2} \end{pmatrix} = \begin{pmatrix} {m_1}_1 & ... & {m_1}_{d_1} \end{pmatrix} M\).
It is advantageous to adopt \(\{{b_1}^1, ..., {b_1}^{d_1}\}\) and \(\{{b_2}^1, ..., {b_2}^{d_2}\}\) instead of \(\{{b_1}_1, ..., {b_1}_{d_1}\}\) and \(\{{b_2}_1, ..., {b_2}_{d_2}\}\), because if we adopted the latter, it would become \(f (m_1)^l = {m_1}^j M^l_j\), which would not be "\(\begin{pmatrix} f (m_1)^1 \\ ... \\ f (m_1)^{d_2} \end{pmatrix} = M \begin{pmatrix} {m_1}^1 \\ ... \\ {m_1}^{d_1} \end{pmatrix}\)", because \(R\) was not necessarily commutative.
But if such a matrix expression is not needed, adopting \(\{{b_1}_1, ..., {b_1}_{d_1}\}\) and \(\{{b_2}_1, ..., {b_2}_{d_2}\}\) is allowed, which makes it \(f (m_1)^l = {m_1}^j M^l_j\).
3: Proof
Whole Strategy: Step 1: see that \(M\) is well-defined and \({m_1}_j M^j_l {b_2}^l\) is uniquely determined; Step 2: suppose that \(f\) is linear; Step 3: see that \(\forall m_1 = {m_1}_j {b_1}^j \in M_1 (f (m_1) = {m_1}_j M^j_l {b_2}^l)\); Step 4: suppose that \(\forall m_1 = {m_1}_j {b_1}^j \in M_1 (f (m_1) = {m_1}_j M^j_l {b_2}^l)\); Step 5: see that \(f\) is linear.
Step 1:
Let us see that \(M\) is well-defined.
\(M\) 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: \(M^j_l {b_2}^l\) is the decomposition of \(f ({b_1}^j)\) with respect to \(B_2\) where \(M^j_l \in R\).
\(M\) is indeed a \(d_1 \times d_2\) \(R\) matrix, because \(j \in \{1, ..., d_1\}\) and \(l \in \{1, ..., d_2\}\): when an \(l\) does not appear in \(M^j_l {b_2}^l\), \(M^j_l := 0\).
The decomposition, \(m_1 = {m_1}_j {b_1}^j\), is unique, by the proposition that for any module with any basis, the components set of any element with respect to the basis is unique: when a \(j\) does not appear in \({m_1}_j {b_1}^j\), \({m_1}_j := 0\).
So, \({m_1}_j M^j_l {b_2}^l\) is uniquely determined for each \(m_1\).
Step 2:
Let us suppose that \(f\) is linear.
Step 3:
Let us see that \(\forall m_1 = {m_1}_j {b_1}^j \in M_1 (f (m_1) = {m_1}_j M^j_l {b_2}^l)\).
\(f (m_1) = f ({m_1}_j {b_1}^j) = {m_1}_j f ({b_1}^j)\), because \(f\) is linear, \(= {m_1}_j M^j_l {b_2}^l\).
Step 4:
Let us suppose that \(\forall m_1 = {m_1}_j {b_1}^j \in M_1 (f (m_1) = {m_1}_j M^j_l {b_2}^l)\).
Step 5:
Let \(m_1 = {m_1}_j {b_1}^j, m'_1 = {m'_1}_j {b_1}^j \in M_1\) and \(r, r' \in R\) be any.
\(f (r m_1 + r' m'_1) = f (r {m_1}_j {b_1}^j + r' {m'_1}_j {b_1}^j) = f ((r {m_1}_j + r' {m'_1}_j) {b_1}^j) = (r {m_1}_j + r' {m'_1}_j) M^j_l {b_2}^l = r ({m_1}_j M^j_l {b_2}^l) + r' ({m'_1}_j M^j_l {b_2}^l) = r f (m_1) + r' f (m'_1)\), which means that \(f\) is linear.
