RE: Rank of a matrix depending on a variable

*To*: mathgroup at smc.vnet.net*Subject*: [mg122013] RE: [mg122000] Rank of a matrix depending on a variable*From*: "David Park" <djmpark at comcast.net>*Date*: Sun, 9 Oct 2011 03:51:29 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <26711112.27416.1318065689598.JavaMail.root@m06>

The problem here is that RowReduce is at once more general and less general than needed. It is more general in incorporating sophisticated numerical techniques, but less general in the ability to confine its operation to a submatrix of a larger matrix but carry the row operations over to all columns. Nevertheless, there is a simple way to do this in regular Mathematica. (A = {{1, 0}, {2, 0}}) // MatrixForm (B = {{a}, {2}}) // MatrixForm Calculate the null space of the transpose of A and then apply this to B. To be consistent all of the resupting elements must be set to zero so this gives a set of equations in the parameters. NullSpace[Transpose@A] %.B {{-2, 1}} {{2 - 2 a}} giving the equation 2 - 2 a == 0. A second method is to use GroebnerBasis. Let the variable symbols be x1 and x2. We then write the equations as polynomials that are equal to zero and find a Groebner basis on x1 (x2 doesn't enter). GroebnerBasis[{x1 - a, 2 x1 - 2}, {x1}] {-1 + a, -1 + x1} giving a == 1 and x == 1 for the solution. This is fine and good if one understands NullSpace and the fact that one has to use the Transpose of the A matrix, or if one is into Groebner bases. Another approach can be used with the Students Linear Equations section of the Presentations application. This allows us to set up a matrix structure representing linear equation, with named rows and columns and dividers. The matrix structure can be displayed in a separate window (or in the notebook) and manipulated step by step. The following does your example. There are routines to compose a matrix structure, operate on it and extract various parts. Unfortunately I can't easily copy the matrix structure display form into the email. A PDF and Mathematica notebook showing the complete solutions will be posted at the archive site Pete Lindsay at the St. Andrews Mathematics department keeps for me, perhaps with a day or so delay. http://home.comcast.net/~djmpark/DrawGraphicsPage.html Here is the code for one method using a null space. 1) Define case1 as a matrix structure with 2 rows and 3 columns. 2) Insert column names, which can actually be variable symbols or constants. 3) Add a divider after the second row. 4) Insert matrix A starting at row 1, column1. 5) Insert matrix B starting at row 1, column 3. 6) Display the structure. << Presentations` case1 = LECreate[2, 3]; LEInsertColumnNames[case1, 1, {x1, x2, 1}] LEColumnDividers[case1, {2}] LEInsertMatrix[case1, {1, 1}, A] LEInsertMatrix[case1, {1, 3}, B] LEPrint[case1] (This displays a matrix with row and column names and dividers.) The following prints the equations: Notice that the expressions are obtained by dotting the row of column names with the row of elements. LERowExpression[case1, #, 1 ;; 2] == LERowExpression[case1, #, 3 ;; 3] & /@ Range[2] // Column[#, Alignment -> "=="] & X1 == a 2 x2 == 2 The following reduces the lhs matrix and carries the row operations over to the rhs without attempting to carry the reduction into the rhs. We then scale the last row to remove a common factor of 2. (The matrix itself is the first part of case1.) LEMakeEchelonForm[case1, {1, 1}, {2, 2}] LERowDividers[case1, {1}] LEScaleRow[case1, 1/2, 2]; case1 = MapAt[Expand, case1, 1]; LEPrint[case1] (Again a display of the reduced matrix structure.) The all zero rows on the lhs generate the null space. For a solution to exist the corresponding rhs entries must be zero, which supplies the condition on the parameter(s). Writing the equations again, we obtain: LERowExpression[case1, #, 1 ;; 2] == LERowExpression[case1, #, 3 ;; 3] & /@ Range[2] Solve[%][[1]] {x1 == a, 0 == 1 - a} {a -> 1, x1 -> 1} Student's Linear Equations has routines for composing matrix structures, routines for row operations on the structure, routines for extracting sub-matrices, row or column expressions in equation form, and routines for displaying the matrix structure in the notebook, or in a separate window where it is dynamically updated. Row, column or element positions can be pasted into the notebook from the displays. By using row and column names the matrices are displayed in a context. There are also routines for displaying rows as reaction equations for use in chemical stoichiometry and operations for basic linear programming. It also has provision for covariant (the usual) and contravariant columns. The capability is useful for didactic purposes and for solving small to medium size problems where one wishes to see the steps. David Park djmpark at comcast.net http://home.comcast.net/~djmpark/ From: Mikel [mailto:ketakopter at gmail.com] Hi, A bit of background for my problem: I'm trying to solve a linear system of equations in matrix form, A.x==B. I have the A matrix, which is square and singular. I also have the B vector, which depends on some parameters (or variables, I'm not sure how to call it). I'm looking for the value of the parameters, so there exists solution to the system. In other words, I want the rank of the augmented matrix AB to be equal to the rank of A. I thought about using the MatrixRank function, but it does not support parameters, it seems. Then I tried RowReduce, but even though in the first example of the help page the result is given with parameters, I only get numbers. Here's an example to try: ============ Clear[a] A = {{1, 0}, {2, 0}}; B = {{a}, {2}}; AB = ArrayFlatten[{{A, B}}]; MatrixRank[A] MatrixRank[AB] RowReduce[AB] a = 1; MatrixRank[A] MatrixRank[AB] RowReduce[AB] ============ As can be seen, the rank does depend on the value of the "a" parameter, but RowReduce just outputs a numeric value. So, the questions are: 1) Why does RowReduce not give a parameter in the result? Is this a bug? 2) Is there any other way to solve the problem, i.e. to get the rank of AB as a function of the "a" variable? I'm using Mathematica 7, Student version, for what is worth.

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**Re: Rank of a matrix depending on a variable**

**Re: Rank of a matrix depending on a variable**