description/proof of that for linear map between vectors spaces and 'vectors spaces - linear morphisms' isomorphism, composition of linear map after isomorphism does not necessarily have 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 knows a definition of direct sum of modules.
Target Context
- The reader will have a description and a proof of the proposition that for a linear map between some vectors spaces and a 'vectors spaces - linear morphisms' isomorphism, the composition of the linear map after the isomorphism does not necessarily have 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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V'_1\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(V_2\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(f_1\): \(: V'_1 \to V_2\), \(\in \{\text{ the linear maps }\}\)
\(V_0\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(V_1\): \(\in \{\text{ the } F \text{ vectors spaces }\}\) such that \(V_1 \subseteq V'_1\)
\(f_0\): \(: V_0 \to V_1\), \(\in \{\text{ the 'vectors spaces - linear morphisms' isomorphisms }\}\)
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Statements:
Not necessarily \(Rank (f_1 \circ f_0) = Rank (f_1)\)
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2: Note
\(V_1\) is not presupposed to be a vectors subspace of \(V'_1\) and \(V_1\) is not presupposed to equal \(V'_1\) (the composition is valid if \(V_1 \subseteq V'_1\)), which are the points.
This proposition is a warning not to jump to the conclusion without checking some things: refer to 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 vectors 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.
3: Proof
Whole Strategy: Step 1: see that \(f_1 \circ f_0\) is not guaranteed to be linear; Step 2: see a counterexample that even if \(V_1\) is a vectors subspace of \(V'_1\), \(Rank (f_1 \circ f_0) \lt Rank (f_1)\).
Step 1:
\(V_1 \subseteq V'_1\) means just that \(V_1\) is a subset of \(V'_1\) not that \(V_1\) is a vectors subspace of \(V'_1\).
For each \(v, v' \in V_0\) and each \(r, r' \in F\), \(f_1 \circ f_0 (r v + r' v') = f_1 (f_0 (r v + r' v')) = f_1 (r f_0 (v) + r' f_0 (v'))\), but the issue here is that \(r f_0 (v) + r' f_0 (v')\) is by the vectors space structure of \(V_1\) not of \(V'_1\), so, \(= r f_1 (f_0 (v)) + r' f_1 (f_0 (v'))\) is not guaranteed.
So, \(f_1 \circ f_0\) is not guaranteed to be linear.
Without being linear, \(Rank (f_1 \circ f_0)\) is not defined.
Step 2:
Let us suppose that \(V_1\) is a vectors subspace of \(V'_1\).
Let \(V'_1 = \mathbb{R}^2\) as the Euclidean vectors space, \(V_2 = \mathbb{R}^2\) as the Euclidean vectors space, \(f_1 = id_{\mathbb{R}^2}: V'_1 \to V_2\) as the identity map, \(V_0 = \mathbb{R} \oplus \{0\}\) where \(\mathbb{R}\) is the Euclidean vectors space, \(V_1 = \mathbb{R} \oplus \{0\}\) where \(\mathbb{R}\) is the Euclidean vectors space, and \(f_0 = id_{\mathbb{R} \oplus \{0\}}: V_0 \to V_1\) as the identity map.
Certainly, \(V_1 \subseteq V'_1\), where \(V_1\) is 1-dimensional, because for example, \(\{(1, 0)\}\) is a basis.
Certainly, \(f_0\) is a 'vectors spaces - linear morphisms' isomorphism.
But \(Ran (f_1) = \mathbb{R}^2\) and \(Rank (f_1) = Dim (Ran (f_1)) = 2\), while \(Ran (f_1 \circ f_0) = \mathbb{R} \otimes \{0\}\) and \(Rank (f_1 \circ f_0) = Dim (Ran (f_1 \circ f_0)) = 1\).
So, \(Rank (f_1 \circ f_0) \lt Rank (f_1)\).
