description/proof of that rank of matrix over field is conserved by multiplying invertible matrices from left and right
Topics
About: matrices space
The table of contents of this article
Starting Context
- The reader knows a definition of field.
- The reader knows a definition of rank of matrix over ring.
- The reader admits the proposition that for any linear map between any finite-dimensional vectors spaces, the rank of the map is the rank of the representative matrix with respect to any bases.
- The reader admits 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.
Target Context
- The reader will have a description and a proof of the proposition that the rank of any matrix over any field is conserved by multiplying any invertible matrices from left and right.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(M\): \(\in \{\text{ the } m \times n F \text{ matrices }\}\)
\(M_1\): \(\in \{\text{ the invertible } m \times m F \text{ matrices }\}\)
\(M_2\): \(\in \{\text{ the invertible } n \times n F \text{ matrices }\}\)
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Statements:
\(Rank (M) = Rank (M_1 M M_2)\)
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2: Proof
Whole Strategy: use the proposition that for any linear map between any finite-dimensional vectors spaces, the rank of the map is the rank of the representative matrix with respect to any bases; Step 1: think of the linear map, \(f: F^n \to F^m\), of which \(M\) is the canonical representative matrix, and see that the rank of \(f\) is \(Rank (M)\); Step 2: think of the linear map, \(f_1: F^m \to F^m\), of which \(M_1\) is the canonical representative matrix, think of the linear map, \(f_2: F^n \to F^n\), of which \(M_2\) is the canonical representative matrix, think of the linear map, \(f': F^n \to F^m\), of which \(M_1 M M_2\) is the canonical representative matrix, and see that the rank of \(f'\) is \(Rank (M_1 M M_2)\); Step 3: see that \(Rank (f') = Rank (f)\).
Step 1:
Let us think of the linear map, \(f: F^n \to F^m\), of which \(M\) is the canonical representative matrix, which means that for each \(v \in F^n\), \(f (v) = M v^t\).
\(Rank (f) = Rank (M)\), by the proposition that for any linear map between any finite-dimensional vectors spaces, the rank of the map is the rank of the representative matrix with respect to any bases.
Step 2:
Let us think of the linear map, \(f_1: F^m \to F^m\), of which \(M_1\) is the canonical representative matrix, which means that for each \(v \in F^m\), \(f_1 (v) = M_1 v^t\).
Let us think of the linear map, \(f_2: F^n \to F^n\), of which \(M_2\) is the canonical representative matrix, which means that for each \(v \in F^n\), \(f_2 (v) = M_2 v^t\).
Let us think of the linear map, \(f': F^n \to F^m\), of which \(M_1 M M_2\) is the canonical representative matrix, which means that for each \(v \in F^n\), \(f' (v) = M_1 M M_2 v^t\).
\(Rank (f') = Rank (M_1 M M_2)\), by the proposition that for any linear map between any finite-dimensional vectors spaces, the rank of the map is the rank of the representative matrix with respect to any bases.
Step 3:
Let us see that \(Rank (f') = Rank (f)\).
\(f' = f_1 \circ f \circ f_2\).
As \(M_1\) is invertible, \(f_1\) is a 'vectors spaces - linear morphisms' isomorphism.
As \(M_2\) is invertible, \(f_2\) is a 'vectors spaces - linear morphisms' isomorphism.
So, \(Rank (f') = Rank (f)\), by 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.
So, \(Rank (M) = Rank (M_1 M M_2)\).
