alexa
Reach Us +44-1522-440391
Classification of Canonical Bases for (nand#8722;1)and#8722;Dimensional Subspaces of n-Dimensional Vector Space | OMICS International
ISSN: 1736-4337
Journal of Generalized Lie Theory and Applications
Make the best use of Scientific Research and information from our 700+ peer reviewed, Open Access Journals that operates with the help of 50,000+ Editorial Board Members and esteemed reviewers and 1000+ Scientific associations in Medical, Clinical, Pharmaceutical, Engineering, Technology and Management Fields.
Meet Inspiring Speakers and Experts at our 3000+ Global Conferenceseries Events with over 600+ Conferences, 1200+ Symposiums and 1200+ Workshops on Medical, Pharma, Engineering, Science, Technology and Business
All submissions of the EM system will be redirected to Online Manuscript Submission System. Authors are requested to submit articles directly to Online Manuscript Submission System of respective journal.

Classification of Canonical Bases for (n−1)−Dimensional Subspaces of n-Dimensional Vector Space

Shtukar U*

Math/Physics Department, North Carolina Central University, USA

*Corresponding Author:
Shtukar U
Associate Professor
Math/Physics Department
North Carolina Central University
1801 Fayetteville Street, Durham
NC 27707, USA
Tel: 919-597-0375
E-mail: [email protected]

Received Date: September 20, 2016; Accepted Date: November 18, 2016; Published Date: November 29, 2016

Citation: Shtukar U (2016) Classification of Canonical Bases for (n-2)-Dimensional Subspaces of n-Dimensional Vector Space. J Generalized Lie Theory Appl 10:245. doi:10.4172/1736-4337.1000245

Copyright: © 2016 Shtukar U. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.

Visit for more related articles at Journal of Generalized Lie Theory and Applications

Abstract

Famous K. Gauss introduced reduced row echelon forms for matrices approximately 200 years ago to solve systems of linear equations but the number of them and their structure has been unknown until 2016 when it was determined at first in the previous article given up to (n−1)×n matrices. The similar method is applied to find reduced row echelon forms for (n−2)×n matrices in this article, and all canonical bases for (n−2)-dimensional subspaces of -dimensional vector space are found also.

Keywords

Vector space; Subspaces; Canonical bases

Introduction

The canonical bases for (n−2)-dimensional subspaces of n-dimensional vector space are introduced in the article, and all nonequivalent of them are classified [1]. Canonical bases for (n−1)- dimensional subspaces of n-dimensional vector space were classified in the previous article [2] of the same author. This new case of (n−2)-dimensional subspaces is interesting to be studied because some n-dimensional Lie algebras haven’t any (n−1)-dimensional subalgebras. For example, in the article [3], it was proved that 6-dimensional Lie algebra of Lorentz group doesn’t have any 5-dimensional subalgebra but this Lie algebra has 4-dimensional subalgebras. We start to introduce the necessary definitions.

Let V be an n-dimensional vector space with its standard basis Equation. Suppose that Equation are (n−2) linearly independent vectors in the space V where,

Equation

The vectors (I) describe all possible bases for any (n−2)-dimensional subspace S of V. This description contains too many arbitrary components; their number is (n−2)×n. Instead of that, we introduce canonical bases with much smaller number of arbitrary components in each of them (maximum 2(n−2)).

Definition 1: Two bases are called equivalent if they generate the same subspace of V, and they are called nonequivalent if they generate two different subspaces of V.

We will associate the following (n−2)×n matrix M with a basis (I)

Equation            (II)

Definition 2: Two matrices are called row equivalent (or just equivalent) if they have the same reduced row echelon form, and they are called nonequivalent if they have different reduced row echelon forms.

About reduced row echelon forms of matrices, see for example [1].

Definition 3: The basis (I) is called canonical if its vectors Equation are the corresponding rows in some reduced row echelon form of the matrix M.

Thus, there is one-to-one correspondence between nonequivalent canonical bases for (n−2)-dimensional subspaces of n-dimensional vector space and nonequivalent reduced row echelon forms for (n−2)×n matrix M of the rank(n−2).

Part I. Basic Examples

Consider two examples of nonequivalent canonical bases for (n−2)-dimensional subspaces of n-dimensional vector spaces where n=4 and n=6.

Ex. 1: Let V be 4-dimensional vector space with its standard basis Equation. Any 2-dimensional subspace S of V can be described as Equation where,

Equation.

This arbitrary basis is equivalent to one and only one canonical basis from the next list:

Equation EquationEquationEquation (5) EquationEquation.

Details of evaluation are omitted because it is similar (but easier) to the evaluation in the example 2. The last canonical bases generate the following 6 matrices associated with them:

Equation.

Ex. 2: Let V be 6-dimensional vector space with its standard basis Equation. Any 4-dimensional subspace S of V can be described as Equation where,

Equation,

Equation     (III)

Start our transforming procedure for the basis Equation to find all possible canonical nonequivalent bases.

A. Let at least one coefficient from a1, b1, c1, d1 in the basis (III) is not zero. Without any loss in the generality, we can suppose that a1 ≠ 0. Perform the linear operation Equation first, and operations Equation after the first one. As a result, the following basis is obtained:

Equation

Remark 1: The first components of vectors Equation are changed as the result of operations performed but all other components of them still have the same notations just for the common convenience. This idea will be used also in all steps of the procedure below.

1. Suppose now that at least one coefficient from b2, c2, d2 at the basis (a) is not zero. Without any loss in generality, let b2 ≠ 0. Perform the linear operations: first Equation, and then Equation. As the result, the following transformed basis is obtained:

Equation       (1)

2. Suppose that at least one coefficient among c3, d3 at the basis (1) is not zero. Again, without any loss in the generality, let c3 ≠ 0. Perform the operation Equation first, and then operations Equation. As the result, the following basis is done.

Equation         (2)

3. Suppose now that the coefficient d4 at the basis (2) is not zero. Perform the operation Equation first, and then operations Equation. As the result, the following canonical basis is obtained:

Equation

If d4=0 then the basis (2) is transformed into the following one:

Equation      (3)

Vector Equation in the basis (3) has at least one nonzero coefficient d5 or d6. Let d5 ≠ 0.

 Perform operation Equation first, and the operations Equation after the first one. As the result, the following canonical basis is obtained:

Equation

If d6 ≠ 0 in the basis (3), then perform operation Equation first, and operations Equation after the first one. We obtain the new canonical basis:

Equation

4. Suppose now that both coefficients c3, d3 at the basis (1) are zero. We have:

Equation       (4)

Consider coefficients c4, d4 in the basis (4). Suppose that at least one of them is not zero. Let c4 ≠ 0. Perform operation Equation first, and perform operations Equation after the first one. The following basis is obtained,

Equation

In the last basis, at least one coefficient d5 or d6 is not zero. Let d5 ≠ 0. Performing operation Equation first, and operations Equation after the first one, we obtain the new canonical basis:

Equation

If d6 0 then doing similarly we obtain the following canonical basis:

Equation

If d6 ≠ 0 in the basis (4), then there will be obtained the bases that are equivalent to (a4) and (a5).

5. Suppose now that both coefficients c4, d4 in the basis (4) are zero. We obtain:

Equation      (5)

In the last basis (5), at least one coefficient among c5, d5 is not zero. If both coefficients d5, f5 are zero, then Equation , and vectors Equation are linearly dependent but it’s impossible for any basis. Let c5≠ 0. Perform the operation Equation first, and operations Equation after the first one. We obtain the following basis:

Equation

In the last basis, d6≠ 0. Perform the operation Equation first, and the operations Equation after the first one. We obtain the new canonical basis:

Equation

At the case when d5≠ 0 in the basis (5), we obtain the same basis (a5).

6. Suppose, in opposition to step 1, that all coefficients b2, c2, d2 in the basis (a) are zero. We obtain:

Equation

Consider coefficients b3, c3, d3 in the last basis. Suppose that at least one of them is not zero. Let b3≠ 0 (without any loss in generality). Perform the operation Equation first, and the operations Equation after the first one. We obtain the following result:

Equation        (6)

7. Suppose now that at least one coefficient from c4, d4 in the basis (6) is not zero. Let c4≠ 0. Perform the operation Equation first, and the operations Equation after the first one. The following basis is obtained:

Equation         (7)

In the basis (7), at least one from the coefficients d5, d6 is not zero.

Let d5 ≠ 0. Perform the operation Equation first, and the operations Equation after the first one. We obtain the new canonical basis:

Equation

If d6 ≠ 0 in the basis (7), we obtain the following canonical basis performing similar steps:

Equation

If d4 0 in the basis (6), we obtain the canonical bases that are equivalent to (a7) and (a8).

8. Suppose now that both coefficients c4, d4 in the basis (6) are zero. We obtain:

Equation      (8)

In the basis (8), at least one coefficient from c5, d5 is not zero. Otherwise, vectors Equation are linearly dependent. Let c5 ≠ 0. Perform the operation Equation first, and the operations Equation after the first one. We obtain the following basis:

Equation

Performing one more obvious step, we obtain the following canonical basis:

Equation

If d5 0 in the basis (8), then we obtain the canonical basis that is equivalent to (a9).

9. Suppose now that all coefficients b2, c2, d2 and b3, c3, d3 in the basis (a) are zero. This is the new case that opposites to the case considered in the step 6. We have:

Equation

In the last basis, at least one coefficient from b4, c4, d4 is not zero. Otherwise, vectors are linearly dependent but it’s impossible for any basis. Let b4 ≠ 0 (without any loss in the generality). Perform the operation Equation first, and the operations Equation after the first one. We obtain the following basis:

Equation      (9)

In the basis (9), at least one coefficient among c5, d5 is not zero. Let c5≠ 0. Perform the operation Equation first, and the operations Equation after the first one. The following basis is obtained:

Equation

The obvious linear operations transform the last basis into the new canonical basis:

Equation

If d5 0 in the basis (9), then the basis that is equivalent to (a10) will be obtained. We have analyzed all possibilities in the situation A.

B. Suppose now that all coefficients a1, b1, c1, d1 are zero in (III). The following basis is obtained:

Equation

1. Consider coefficients a2, b2, c2, d2 in the basis (b). Suppose now that at least one coefficient among a2, b2, c2, d2 is not zero. Without any loss in generality, let a2 ≠ 0. Perform next linear operations: Equation first, and Equation then. As the result, the next transformed basis appears:

Equation      (1)

2. Suppose now that at least one coefficient from b3, c3, d3 in the basis (1) is not zero. Without any loss in generality, let b3 0. Perform next linear operations: Equation first, and Equation then. As the result, the next transformed basis appears:

Equation        (2)

3. Consider coefficients c4, d4 in the basis (2). Suppose that at least one of them is not zero. Let c4 ≠ 0. Perform operation Equation first, and operations Equation after the first one. We obtain the following basis:

Equation     (3)

The vector Equation in the basis (3) has at least one non zero coefficient in the last basis. If d5 ≠ 0 then performing operations Equation , we obtain the new canonical basis:

Equation

If d6 ≠ 0 then we obtain one more new canonical basis:

Equation

The assumption d4 0 brings the same canonical bases (b1) and (b2).

4. Suppose now that both coefficients c4, d4 are zero in the basis (2). We have:

Equation       (4)

In the basis (4), at least one coefficient from c5, d5 is not zero. Let c5 ≠ 0. Perform operation Equation first, and operations Equation after the first one. We obtain the next basis:

Equation

It is easy to transform the last basis into the next canonical basis:

Equation

If we suppose that d5 ≠ 0 in the basis (4), the same canonical basis (b3) will be done.

5. Suppose now that all coefficients b3, c3, d3 in the basis (1) are zero. We obtain:

Equation        (5)

In the basis (5), at least one coefficient among b4, c4, d4 is not zero. Otherwise, vectors Equation are linearly dependent but it’s impossible. Let b4 ≠ 0. Perform next linear operations: Equation first, and Equation then. As the result, the next transformed basis appears:

Equation

If c4 ≠ 0 or d4 0, then we obtain bases that are equivalent to the last basis. At least one coefficient among c5, d5 at the last basis is not zero. Let c5 ≠ 0. Perform operation Equation first, and operations Equation after the first one. We obtain:

Equation

The last basis can be transformed immediately into the next canonical basis:

Equation

If we suppose that d5 0 instead c5 ≠ 0, we’ll obtain the same canonical basis (b4).

6. Consider coefficients a2, b2, c2, d2 in the basis (b). Suppose now (in opposition to the step 1) that all coefficients a2, b2, c2, d2 are zero. We have the basis:

Equation       (6)

Consider coefficients a3, b3, c3, d3. At least one of them is not zero. Otherwise, vectors Equation are linearly dependent but it’s impossible for any basis. Let a3 ≠ 0 (without any loss in generality). Perform the operation Equation first, and the operations Equation after the first one. We obtain the following basis:

Equation

Consider coefficients b4, c4, d4 in the last basis. At least one of them is not zero. Otherwise, vectors Equation are linearly dependent but it’s impossible. Let b4 ≠ 0. Perform the operation Equation first, and the operations Equation after the first one. We obtain:

Equation

Continue this procedure; we will obtain the following canonical basis at the end:

Equation

All other subcases in the step 6 give the same basis (b5).

The total list of all canonical bases that are found at the situations A and B is done here:

Equation

Equation

Compare these canonical bases to determine nonequivalent among them. If d5 ≠ 0 in the basis (a3) then this basis is equivalent to the basis (a2), so d5=0 in (a3). Similarly, if d5 ≠ 0 in the basis (a5) then this basis is equivalent to the basis (a4), so d5=0 in (a5). Again, if d5 ≠ 0 in the basis (a8) then this basis is equivalent to the basis (a7), so d5=0 in (a8). If d5 ≠ 0 in the basis (b2) then this basis is equivalent to the basis (b1), so d5=0 in (b2). The final list of nonequivalent canonical bases is:

Equation

The following 15 matrices are associated with the canonical bases above:

Equation

Equation

All these matrices are not equivalent.

Part II. General Case:

The following statement concerning (n−2)×n matrices M of the type (II) is true.

Theorem 1: Let M be a (n−2)×n matrix (II) of the rank (n−2) where n ≥ 4. This matrix is row equivalent to one and only one of the Equation following matrices:

Equation

Equation

All matrices above are not equivalent between them.

Proof: We will use the mathematical induction method with respect to the dimension n. This statement is correct in the cases n=4 and n=6 according Examples 1 and 2. Suppose that the statement is true for arbitrary n≥4, and prove it for the dimension (n+1). Let M be a matrix of the size (n−1)× (n+1):

Equation

Consider the (n−2)× n submatrix M′ located in the upper left corner of the matrix M. According the assumption, this submatrix can be transformed into one of the matrices listed in this statement. We will substitute submatrix M′ by the corresponding matrix, and then transform the special matrix M into reduced rom echelon form. The standard linear operations with rows (vectors) will be utilized: (a) interchange any two rows, (b) multiply any row by a nonzero constant, (c) add a multiple of some row to another row.

1. At the first case, we have:

Equation

Perform linear transformations Equation The result of the operations is the following matrix:

Equation

At least one components among Equation is not zero but all other components of the (n−1) row are zero. Let Equation. Perform the operation Equation first, and the operationsEquation after the first one. We obtain:

Equation

It is the matrix of the first type from the list above as we need. Let Equation. Perform the operation Equation first, and the operations Equation after the first one. We obtain:

Equation

The last matrix is of the second type matrix from the list as we need. Let Equation and Equation. Perform the operation Equation first, and the operationsEquation after the first one. We obtain:

Equation

The last matrix is of the (n−1) type matrix from the list above. The statement is proved for the 1st case.

2. At the second case, we have:

Equation

Perform linear transformations Equation and Equation. The result of the operations is the following matrix:

Equation

At least one components among Equation is not zero but all other components of the (n−1) row are zero. Let an−1,n−2 ≠ 0. Perform the operation Equation first, and the operations Equation after the first one. We obtain:

Equation

If interchange rows Equation in the last matrix, we obtain the matrix of the first type as we need. Let an−1,n−2=0 and an−1,n≠0. Perform the operation Equation first, and the operations Equation after the first one. We obtain:

Equation

It is the matrix of the 3rd type that follows the matrix of the second type in the list. Let Equation Perform the operation Equation first, and the operations Equation after the first one. We obtain:

Equation

It is the matrix of the type (n) in the list above. The statement is proved for the 2nd case. All the next cases are similar to the cases (1) and (2) but we consider some of them.

Case (n). At this case we have:

Equation

Perform linear transformations Equation and Equation. The result of the operations is the following matrix:

Equation

At least one components among Equation is not zero but all other components of the (n−1) row are zero. Let Equation. Perform the operation Equation first, and the operations Equation after the first one. We obtain:

Equation

If we interchange the last 3 rows of this matrix, we obtain the matrix of the 2nd type from the list as we need. Let Equation, andEquation. Perform the operation Equation first, and the operations Equation after the first one. We obtain:

Equation

If we interchange the rows (n-1) and (n-2) in this matrix, we obtain the matrix of the 3rd type from the list as we need. Let Equation, Equation. Perform the operationEquation first, and the operations Equation after the first one. We obtain:

Equation

It is the matrix of the (2n-3) type from the list. The case (n) is proved.

Case Equation. At this case, we have the following matrix of (n-1)× (n+1) size.

Equation

Perform the operations Equation. We obtain:

Equation

At least one component among Equation is not zero. LetEquation. Perform the operation Equation, and remove the new last row into the first position. We obtain:

Equation

It is the matrix of the 2nd type from the list as we need. Let Equation and Equation in the previous matrix. Perform the operation Equation first, and then remove the new last row into the first position. We obtain:

Equation

It is the matrix of the type Equation from the list as we need. Let Equation in the previous matrix. Perform the operation Equation first, and the operationsEquation after the first one. We have:

Equation

It is the matrix of the Equation type from the list as we need. This case is proved, and the total proof is done.

Remark 2: Of cause, the list of matrices in Theorem 1 doesn’t contain all of them. But any missed matrix can be restored using Ladder Principle. For each subsequence of matrices (between (;) sings) in the list, imagine the ladder from the lower right corner to the upper left corner. Take the left most columns with arbitrary components, and make 1 step up along the ladder bringing this column up and to the left of the previous position. Fix elements 0 and only one element 1 at the corresponding positions in the released column. The next matrix from the list will be done.

As an obvious consequence of Theorem 1, we obtain the following statement.

Theorem 2: Each basis for (n−2)-dimensional subspaces of a n-dimensional vector space (n4) is equivalent to one and only one canonical basis from the following list.

Equation

Equation

Conclusion

Results of this article are ready to be used at any research concerning subalgebras and ideals of noncommutative algebras. Classification of canonical bases for (n−2)-dimensional subspaces is very effective to study reductive subalgebras and reductive pairs of any n-dimensional Lie algebra.

References

Select your language of interest to view the total content in your interested language
Post your comment

Share This Article

Relevant Topics

Article Usage

  • Total views: 8657
  • [From(publication date):
    December-2016 - Sep 15, 2019]
  • Breakdown by view type
  • HTML page views : 8536
  • PDF downloads : 121
Top