description/proof of that for linear map between vectors spaces, 'vectors spaces - linear morphisms' isomorphism onto domain of linear map, and 'vectors spaces - linear morphisms' isomorphism from superspace of codomain of linear map, composition of linear map after 1st isomorphism and before 2nd isomorphism has rank of linear map
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of %field name% vectors space.
- The reader knows a definition of %category name% isomorphism.
- The reader knows a definition of rank of linear map between vectors spaces.
- The reader admits the proposition that the range of any linear map between any vectors spaces is a vectors subspace of the codomain.
- The reader admits the proposition that for any 'vectors spaces - linear morphisms' isomorphism, its restriction on any subspace domain and the corresponding range codomain is a 'vectors spaces - linear morphisms' isomorphism.
- The reader admits the proposition that for any 'vectors spaces - linear morphisms' isomorphism, the image of any linearly independent subset or any basis of the domain is linearly independent or a basis on the codomain.
Target Context
- The reader will have a description and a proof of the proposition that for any linear map between any vectors spaces, any 'vectors spaces - linear morphisms' isomorphism onto the domain of the linear map, and any 'vectors spaces - linear morphisms' isomorphism from any superspace of the codomain of the linear map, the composition of the linear map after the 1st isomorphism and before the 2nd isomorphism has the rank of the linear map.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V_1\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(V_2\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(f_1\): \(: V_1 \to V_2\), \(\in \{\text{ the linear maps }\}\)
\(V_0\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(f_0\): \(: V_0 \to V_1\), \(\in \{\text{ the 'vectors spaces - linear morphisms' isomorphisms }\}\)
\(V'_2\): \(\in \{\text{ the } F \text{ vectors spaces }\}\), such that \(V_2 \in \{\text{ the vectors subspaces of } V'_2\}\)
\(V_3\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(f_2\): \(: V'_2 \to V_3\), \(\in \{\text{ the 'vectors spaces - linear morphisms' isomorphisms }\}\)
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Statements:
\(Rank (f_2 \circ f_1 \circ f_0) = Rank (f_1)\)
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2: Note
The codomain of \(f_0\) is \(V_1\) with the vectors space structure of \(V_1\) and \(V_2\) is a vectors subspace of \(V'_2\) (not just a subset), which are the points.
Refer to the proposition that for any linear map between any vectors spaces and any 'vectors spaces - linear morphisms' isomorphism, the composition of the linear map after the isomorphism does not necessarily have the rank of the linear map.
3: Proof
Whole Strategy: Step 1: see that \(f_2 \circ f_1 \circ f_0\) is linear; Step 2: see that \(Ran (f_1 \circ f_0) = Ran (f_1)\); Step 3: see that \(f_2 \vert_{Ran (f_1)}: Ran (f_1) \to f_2 (Ran (f_1))\) is a 'vectors spaces - linear morphisms' isomorphism, see that \(f_2 \circ f_1 \circ f_0 = f_2 \vert_{Ran (f_1)} \circ f_1 \circ f_0\), and see that \(Dim (Ran (f_2 \circ f_1 \circ f_0)) = Dim (Ran (f_1 \circ f_0))\).
Step 1:
\(f_2 \circ f_1 \circ f_0\) is linear, which is because the codomain of \(f_0\) is the domain of \(f_1\) (more generally, a vectors subspace of the domain of \(f_1\)) and the codomain of \(f_1\) is a vectors subspace of the domain of \(f_2\): just \(V_1 \subseteq V_1\) and \(V_2 \subseteq V'_2\) sets-wise does not guarantee the linearity.
We needed to check the linearity, because otherwise, \(Rank (f_2 \circ f_1 \circ f_0)\) would not be defined.
Step 2:
Let us see that \(Ran (f_1 \circ f_0) = Ran (f_1)\).
For each \(v \in Ran (f_1 \circ f_0)\), \(v = f_1 (f_0 (v_0))\) for a \(v_0 \in V_0\), but \(v_1 := f_0 (v_0) \in V_1\) and \(v = f_1 (v_1) \in Ran (f_1)\).
For each \(v \in Ran (f_1)\), \(v = f_1 (v_1)\) for a \(v_1 \in V_1\), but as \(f_0\) is surjective, there is a \(v_0 \in V_0\) such that \(v_1 = f_0 (v_0)\), so, \(v = f_1 (v_1) = f_1 (f_0 (v_0)) \in Ran (f_1 \circ f_0)\).
That implies that \(Rank (f_1 \circ f_0) = Dim (Ran (f_1 \circ f_0)) = Dim (Ran (f_1)) = Rank (f_1)\).
Step 3:
\(Ran (f_1)\) is a vectors subspace of \(V_2\), by the proposition that the range of any linear map between any vectors spaces is a vectors subspace of the codomain.
\(Ran (f_1)\) is a vectors subspace of \(V'_2\), because \(V_2\) is a vectors subspace of \(V'_2\).
\(f_2 \vert_{Ran (f_1)}: Ran (f_1) \to f_2 (Ran (f_1))\) is a 'vectors spaces - linear morphisms' isomorphism, by the proposition that for any 'vectors spaces - linear morphisms' isomorphism, its restriction on any subspace domain and the corresponding range codomain is a 'vectors spaces - linear morphisms' isomorphism.
\(f_2 \circ f_1 \circ f_0 = f_2 \vert_{Ran (f_1)} \circ f_1 \circ f_0\).
So, \(Dim (Ran (f_2 \circ f_1 \circ f_0)) = Dim (Ran (f_2 \vert_{Ran (f_1)} \circ f_1 \circ f_0))\).
But as \(Ran (f_1 \circ f_0) = Ran (f_1)\), \(Ran (f_2 \vert_{Ran (f_1)} \circ f_1 \circ f_0) = Ran (f_2 \vert_{Ran (f_1)})\), so, \(Dim (Ran (f_2 \vert_{Ran (f_1)} \circ f_1 \circ f_0)) = Dim (Ran (f_2 \vert_{Ran (f_1)}))\), but \(Dim (Ran (f_2 \vert_{Ran (f_1)})) = Dim (Ran (f_1))\), by the proposition that for any 'vectors spaces - linear morphisms' isomorphism, the image of any linearly independent subset or any basis of the domain is linearly independent or a basis on the codomain, so, \(Rank (f_2 \circ f_1 \circ f_0) = Dim (Ran (f_2 \circ f_1 \circ f_0)) = Dim (Ran (f_2 \vert_{Ran (f_1)} \circ f_1 \circ f_0)) = Dim (Ran (f_1)) = Rank (f_1)\).
