description/proof of that for linear map between finite-dimensional vectors spaces, rank of map is rank of representative matrix w.r.t. bases
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of rank of linear map between vectors spaces.
- The reader knows a definition of rank of matrix over ring.
- The reader admits the proposition that for any finite-dimensional vectors space and any basis, the vectors space is 'vectors spaces - linear morphisms' isomorphic to the components vectors space with respect to the basis.
- 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.
- The reader admits the proposition that for any finite-dimensional columns or rows module and any linearly independent subset, any expansion of the subset into any larger-dimensional columns or rows module is linearly independent.
- The reader admits the proposition that for any finite-dimensional columns or rows vectors space and any linearly independent subset, the subset can be shrunk into a number-of-elements-dimensional columns or rows vectors space by choosing some components.
- The reader admits the proposition that for any vectors space, any finite generator can be reduced to be a basis.
- The reader admits the proposition that for any vectors space, the bases have the same cardinality.
- The reader admits the proposition that the determinant of any square matrix over any field is nonzero if and only if the set of the columns or the rows is linearly independent.
Target Context
- The reader will have a description and a proof of 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.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V_1\): \(\in \{\text{ the } d_1 \text{ -dimensional } F \text{ vectors spaces }\}\)
\(V_2\): \(\in \{\text{ the } d_2 \text{ -dimensional } F \text{ vectors spaces }\}\)
\(f\): \(: V_1 \to V_2\), \(\in \{\text{ the linear maps }\}\)
\(B_1\): \(\in \{\text{ the bases for } V_1\}\)
\(B_2\): \(\in \{\text{ the bases for } V_2\}\)
\(M\): \(= \text{ the representative matrix for } f\) with respect to \(B_1\) and \(B_2\)
//
Statements:
\(Rank (f) = Rank (M)\)
//
2: Proof
Whole Strategy: Step 1: see that \(V_2\) is 'vectors spaces - linear morphisms' isomorphic to the components vectors space and see that the rank of \(f\) is the rank of the linear map by \(M\); Step 2: suppose that \(Rank (M) = d\) and the top-left \(d \times d\) submatrix is determinant nonzero; Step 3: see that \(Rank (f) = d\); Step 4: see that when \(Rank (f) = d\), \(Rank (M) = d\).
Step 1:
\(V_2\) is 'vectors spaces - linear morphisms' isomorphic to the components vectors space, by the proposition that for any finite-dimensional vectors space and any basis, the vectors space is 'vectors spaces - linear morphisms' isomorphic to the components vectors space with respect to the basis.
That means that any subspace of \(V_2\) has the same dimension with the corresponding subspace of the components vectors space, 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: any basis of the subspace of \(V_2\) is mapped to a basis of the corresponding subspace of the components vectors space: the restriction of the 'vectors spaces - linear morphisms' isomorphism is obviously a 'vectors spaces - linear morphisms' isomorphism.
So, \(Rank (f)\) is the dimension of the range of the components vectors spaces linear map by \(M\).
Step 2:
Let us suppose that \(Rank (M) = d\).
Let the top-left \(d \times d\) submatrix, \(N\), be determinant nonzero, which is possible by reordering \(B_1\) and \(B_2\).
Step 3:
Let us take the set of \(d\)-dimensional column vectors, \(\{(1, 0, ..., 0)^t, ..., (0, ..., 0, 1)^t\}\), which is obviously linearly independent.
\(\{N (1, 0, ..., 0)^t, ..., N (0, ..., 0, 1)^t\}\), which is a set of \(d\)-dimensional column vectors, is linearly independent, 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: the linear map by \(N\) is a 'vectors spaces - linear morphisms' isomorphism, because \(det N \neq 0\).
Let us think of the \(d_1\)-dimensional expansion of \(\{(1, 0, ..., 0)^t, ..., (0, ..., 0, 1)^t\}\), \(\{(1, 0, ..., 0, 0, ..., 0)^t, ..., (0, ..., 0, 1, 0, ..., 0)^t\}\) and \(\{M (1, 0, ..., 0, 0, ..., 0)^t, ..., M (0, ..., 0, 1, 0, ..., 0)^t\}\).
\(\{M (1, 0, ..., 0, 0, ..., 0)^t, ..., M (0, ..., 0, 1, 0, ..., 0)^t\}\) is a \(d_2\)-dimensional expansion of \(\{N (1, 0, ..., 0)^t, ..., N (0, ..., 0, 1)^t\}\), because the 1st \(d\) components are not changed, and so, is linearly independent, by the proposition that for any finite-dimensional columns or rows module and any linearly independent subset, any expansion of the subset into any larger-dimensional columns or rows module is linearly independent.
So, the range of the map by \(M\) is equal to or larger than \(d\)-dimensional.
Let us suppose that the range was \(d'\)-dimensional where \(d \lt d'\).
There would be a basis for the range, and some \(d'\) components could be chosen such that the shrunk basis was still linearly independent, by the proposition that for any finite-dimensional columns or rows vectors space and any linearly independent subset, the subset can be shrunk into a number-of-elements-dimensional columns or rows vectors space by choosing some components.
Let us move the chosen \(d'\) components to the top \(d'\) rows by reordering \(B_2\), calling the result \(M\) hereafter.
That would mean that the shrunk range (the projection to the 1st \(d'\) components) was \(d'\)-dimensional.
Let \(M'\) be the top \(d' \times d_1\) submatrix of \(M\).
\(\{M' \begin{pmatrix} 1 \\ 0 \\ ... \\ 0 \end{pmatrix}, ..., M' \begin{pmatrix} 0 \\ ... \\ 0 \\ 1 \end{pmatrix}\}\) would span (generate) the shrunk range, obviously.
But it would be nothing but the columns of \(M'\).
So, some \(d'\) columns of \(M'\) could be chosen to be a basis for the shrunk range, by the proposition that for any vectors space, any finite generator can be reduced to be a basis and the proposition that for any vectors space, the bases have the same cardinality.
Let us move the corresponding \(d'\) columns of \(M\) to the left by reordering \(B_1\), calling the result \(M\) hereafter.
Then, the top-left \(d' \times d'\) submatrix would be determinant nonzero, by the proposition that the determinant of any square matrix over any field is nonzero if and only if the set of the columns or the rows is linearly independent, so, \(d \lt d' \le Rank (M)\), a contradiction.
So, the range is \(d\)-dimensional.
So, the rank of the linear map by \(M\) is \(d\), so, \(Rank (f) = d\).
Step 4:
Let us suppose that \(Rank (f) = d\).
If \(Rank (M) = d'\), \(Rank (f) = d'\), by Step 3, so, \(d = d'\).
