Re: Rank of a matrix depending on a
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- Subject: [mg122041] Re: Rank of a matrix depending on a
- From: DrMajorBob <btreat1 at austin.rr.com>
- Date: Mon, 10 Oct 2011 04:28:04 -0400 (EDT)
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- References: <26711112.27416.1318065689598.JavaMail.root@m06>
- Reply-to: drmajorbob at yahoo.com
Brilliant!
As usual, David's package allows us to "eschew obfuscation".
Bobby
On Sun, 09 Oct 2011 02:51:29 -0500, David Park <djmpark at comcast.net> wrote:
> 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.
>
>
--
DrMajorBob at yahoo.com