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 \(d_2 \times d_1\) 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 \(d_2 \times d_1\) 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, \(B_2 = \{b_{2, 1}, ..., b_{2, d_2}\}\)
\(f\): \(: M_1 \to M_2\)
\(M\): \(\in \{\text{ the } d_2 \times d_1 R \text{ matrices }\}\), such that \(f (b_{1, j}) = M^l_j b_{2, l}\)
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Statements:
\(f \in \{\text{ the linear maps }\}\)
\(\implies\)
\(\forall v_1 = {v_1}^j b_{1, j} \in M_1 (f (v_1) = {v_1}^j M^l_j b_{2, l})\)
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2: Note
\(\forall v_1 = {v_1}^j b_{1, j} \in M_1 (f (v_1) = {v_1}^j M^l_j b_{2, l})\) means nothing but "\(f\) is represented by \(M\)", because \(M\) determines \(f\), because the decomposition, \(v^j b_{1, j}\), 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 \(R\) is commutative, it means that \(\begin{pmatrix} {v_2}^1 \\ ... \\ {v_2}^{d_2} \end{pmatrix} = M \begin{pmatrix} {v_1}^1 \\ ... \\ {v_1}^{d_1} \end{pmatrix}\) where \(v_2 := f (v_1) = {v_2}^l b_{2, l}\); when \(R\) is not commutative, \({v_1}^j M^l_j b_{2, l}\) cannot be expressed so, because \({v_1}^j M^l_j \neq M^l_j {v_1}^j\) in general while \(M \begin{pmatrix} {v_1}^1 \\ ... \\ {v_1}^{d_1} \end{pmatrix}\) produces the latter.
3: Proof
Whole Strategy: Step 1: see that \(M\) is well-defined and \({v_1}^j M^l_j b_{2, l}\) is uniquely determined; Step 2: suppose that \(f\) is linear; Step 3: see that \(\forall v_1 = {v_1}^j b_{1, j} \in M_1 (f (v_1) = {v_1}^j M^l_j b_{2, l})\); Step 4: suppose that \(\forall v_1 = {v_1}^j b_{1, j} \in M_1 (f (v_1) = {v_1}^j M^l_j b_{2, l})\); 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: \(M^l_j b_{2, l}\) is the decomposition of \(f (b_{1, j})\) with respect to \(B_2\) where \(M^l_j \in R\).
\(M\) is indeed a \(d_2 \times d_1\) \(R\) matrix, because \(l \in \{1, ..., d_2\}\) and \(j \in \{1, ..., d_1\}\): when an \(l\) does not appear in \(M^l_j b_{2, l}\), \(M^l_j := 0\).
The decomposition, \(v_1 = {v_1}^j b_{1, j}\), 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 \({v_1}^j b_{1, j}\), \({v_1}^j := 0\).
So, \({v_1}^j M^l_j b_{2, l}\) is uniquely determined.
Step 2:
Let us suppose that \(f\) is linear.
Step 3:
Let us see that \(\forall v_1 = {v_1}^j b_{1, j} \in M_1 (f (v_1) = {v_1}^j M^l_j b_{2, l})\).
\(f (v_1) = f ({v_1}^j b_{1, j}) = {v_1}^j f (b_{1, j})\), because \(f\) is linear, \(= {v_1}^j M^l_j b_{2, l}\).
Step 4:
Let us suppose that \(\forall v_1 = {v_1}^j b_{1, j} \in M_1 (f (v_1) = {v_1}^j M^l_j b_{2, l})\).
Let \(v_1 = {v_1}^j b_{1, j}, v'_1 = {v'_1}^j b_{1, j} \in M_1\) and \(r, r' \in R\) be any.
\(f (r v_1 + r' v'_1) = f (r {v_1}^j b_{1, j} + r' {v'_1}^j b_{1, j}) = f ((r {v_1}^j + r' {v'_1}^j) b_{1, j}) = (r {v_1}^j + r' {v'_1}^j) M^l_j b_{2, l} = r ({v_1}^j M^l_j b_{2, l}) + r' ({v'_1}^j M^l_j b_{2, l}) = r f (v_1) + r' f (v'_1)\), which means that \(f\) is linear.
