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Consider the following code, which just generates two "identical" matrices, differing only in their requested storage type, and then does some simple manipulations.

restart;
#
# Define matrix using sparse storage
#
   testM:= Matrix( 40,40,
                           (i,j)->`if`(j>=i,1,0),fill=0,
                           storage=sparse
                        ):
#
# Define identical(?) matrix with rectangular storage
#
   nm:= Matrix( 40,40,
                        testM,
                        storage=rectangular
                     ):
#
# Define procedure to return some matrix properties
#
   matData:= proc( myMat::Matrix)
                            return op(3, myMat)[2], # check storage type
                                      myMat[5, 1..-1], # get 5-th row
                                      add(myMat[5, 1..-1]); # add elements in 5-th row
                    end proc:
#
# Get properies of the two matrices - should be identical
# but check result of adding elements in the 5-th row
#
    matData(testM);
    matData(nm);

The matData procedure ought to produce the same results for the two matrices, with the exception of the storrage type. But the 'add()' command does not. The 'myMat[5, 1..-1]' command produces the same vector, the 5-th row - but stick an add() wrapper around it and all hell breaks loose.

Is this a bug or am I missing something?

Suggestions such as avoiding sparse data storage are not really acceptable: the above is a much simplified version of my original problem where I was using graph theory to play with a "cost function" and (with G a graph) the command,

WeightMatrix(MinimalSpanningTree(G))

returned a sparse-storage matrix - and I didn't notice. There appears to be no option on the WeightMatrix() command to control the storage tyoe of the returned matrix. Result was that all subsequent code based on slicing/dicing/and particularly 'add()ing' sub-blocks of this weight matrix fell apart

Don't get me wrong: I can sort of accept that the weight matrix of minimal spanning tree would (hopefully) be mainly zeros so sparse-storage might be a good default option but I don't see why the results of a command such as

add(myMat[5, 1..-1])

should vary depending on the internal storage used for the matrix, particularly when I have no control over the storage type being adopted

 

Maple 18.02 on windows. A 4 by 4 matrix, does not display on the screen in nice formating when it has too many elements to fit current screen. But I'd like it to be displayed in 2D just like all the other 4 by 4 matrices and then use the horizontal scroll bar if needed to see the full matrix. Is this possible?

------------------------------------

restart;
z:=theta__1:
T01:=Matrix([
[cos(z),   0,   sin(z),   L*cos(z)],
[-sin(z),  0,  -cos(z),   L*sin(z)],
[0,         1,  0,          0],
[0,         0,  0,          1]]):

z:=theta__2:
T12:=Matrix([
[cos(z),    0,   -sin(z),   L*cos(z)],
[sin(z),    0,    cos(z),   L*sin(z)],
[0,         1,  0,          0],
[0,         0,  0,          1]]):

z:=theta__3:
T23:=Matrix([
[cos(z),    0,   -sin(z),   L*cos(z)],
[sin(z),    0,    cos(z),   L*sin(z)],
[0,         -1,  0,          0],
[0,         0,  0,          1]]):

T02   := T01.T12;
T03   := T02.T23;
LinearAlgebra[Dimension](T03);
------------------------------------------------

T02 above displays in 2D fine. But T03 does not on standard 100% zoom on my monitor. Screen shot:

When I changed the zoom to 50%, now it did format ok on the screen:

May be I need a way to activate the horizontal screel bar? I really do not want to keep changing zoom each time I want to see a larger matrix. All the matrices are 4 by 4, but some of them can end up with many terms in each entry.

Hi,

for my simulation I have to calculate several gradients and jacobian matrices. The equations are quite complex and with my current setup hard to read.

 

Here is some exampel code:

restart;
with(VectorCalculus);
SetCoordinates(cartesian[x, y, z]);
alias(u = u(t, x, y, z), v = v(t, x, y, z), w = w(t, x, y, z)); alias(eta = eta(t, x, y, z));
U := VectorField(`<,>`(u, v, w));
Divergence(U);
Jacobian(U);
Diff(U, t);

The Divergence operator gives me a very compact result:

But Jaobian and Diff look like:

 

To achive a better readability I want to do two things (if possible):

1) hide the independet variables (t,x,y,z) in the result of Jacobian, Diff

2) display the result of Diff in a row vector (3x1) instead

 

Is this possible?

Thanks in advance for your help

 

 

I am trying to learn the <> notation to enter matrices and vectors. But I find this page very confusing

http://www.maplesoft.com/support/help/maple/view.aspx?path=examples%2FLA_Syntax_Shortcuts

it says:

but we see clearly the vertical bars are used to separate columns.

Isn't a column the thing that goes from the top to bottom and not from left to right in Maple LinearAlgebra?

 

Hi all

Can anybody suggest an algorithm allowing to detect, that two matrices of the same size can be obtained each from other by permutations of rows and columns? Maybe, such an algorithm there exists in LinearAlgebra package?

is it possible to assume element of matrix 0 or 1

how to write?

after write this, is it possible to display possible matrices which each element is 0 or 1

with(LinearAlgebra):
GetRing := proc(sol)
ringequation := 0;
mono1 := 0;
for j from 1 to 3 do
mono1 := 1;
for i from 1 to nops(sol[1][j]) do
mono1 := mono1*op(i, sol[1][j]);
od:
ringequation := ringequation + mono1;
od:
return ringequation;
end proc;
M := Matrix([[a1,a2,a3],[a4,a5,a6],[a7,a8,a9]]);
sys := a*b+a = GetRing(MatrixMatrixMultiply(Matrix([[a,b,c]]), M));
solve(sys);

This application creates DNG matrices by optimizing Delta E from a raw photo of x-rites color checker. The color temperature for the photograph is also estimated.  Inputs are raw data from RawDigger and generic camera color response from DXO Mark.

Initialization

   

NULL

NULL

NULL

NULL

NULL

XYZoptical to RGB to XYZdata

 

 

Sr,g,b is the relative spectral transmittance of the filter array not selectivity for XY or Z of a given color.

Pulling Sr,g,b out of the integral assumes they are scalars. For example Sr attenuates X, Y and Z by the same amount.

Raw Balance is not White Point Adaptation.

The transmission loss of Red and Blue pixels relative to green is compensated by D=inverse(S). The relation to incident chromaticity, xy is unchanged as S.D=1.

(See Bruce Lindbloom; "Spectrum to XYZ" and "RGB/XYZ Matrices" also, Marcel Patek; "Transformation of RGB Primaries")

 

 

X = (Int(I*xbar*S, lambda))/N:

Y = (Int(I*ybar*S, lambda))/N:

Z = (Int(I*zbar*S, lambda))/N:

N = Int(I*ybar, lambda):

• 

XYZ to RGB

(Vector(3, {(1) = R_Tbb, (2) = G_Tbb, (3) = B_Tbb})) = (Matrix(3, 3, {(1, 1) = XR*Sr, (1, 2) = YR*Sr, (1, 3) = ZR*Sr, (2, 1) = XG*Sg, (2, 2) = YG*Sg, (2, 3) = ZG*Sg, (3, 1) = XB*Sb, (3, 2) = YB*Sb, (3, 3) = ZB*Sb})).(Vector(3, {(1) = X_Tbb, (2) = Y_Tbb, (3) = Z_Tbb}))

NULL

(Vector(3, {(1) = R_Tbb, (2) = G_Tbb, (3) = B_Tbb})) = (Matrix(3, 3, {(1, 1) = Sr, (1, 2) = 0, (1, 3) = 0, (2, 1) = 0, (2, 2) = Sg, (2, 3) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = Sb})).(Matrix(3, 3, {(1, 1) = XR, (1, 2) = YR, (1, 3) = ZR, (2, 1) = XG, (2, 2) = YG, (2, 3) = ZG, (3, 1) = XB, (3, 2) = YB, (3, 3) = ZB})).(Vector(3, {(1) = X_Tbb, (2) = Y_Tbb, (3) = Z_Tbb}))

 

Camera_Neutral = (Matrix(3, 3, {(1, 1) = Sr, (1, 2) = 0, (1, 3) = 0, (2, 1) = 0, (2, 2) = Sg, (2, 3) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = Sb})).(Matrix(3, 3, {(1, 1) = XR, (1, 2) = YR, (1, 3) = ZR, (2, 1) = XG, (2, 2) = YG, (2, 3) = ZG, (3, 1) = XB, (3, 2) = YB, (3, 3) = ZB})).(Vector(3, {(1) = X_wht, (2) = Y_wht, (3) = Z_wht}))

NULL

NULL

NULL

• 

RGB to XYZ (The extra step of adaptation to D50 is included below)

 

(Vector(3, {(1) = X_D50, (2) = Y_D50, (3) = Z_D50})) = (Matrix(3, 3, {(1, 1) = XTbbtoXD50, (1, 2) = YTbbtoXD50, (1, 3) = ZTbbtoXD50, (2, 1) = XTbbtoYD50, (2, 2) = YTbbtoYD50, (2, 3) = ZTbbtoYD50, (3, 1) = XTbbtoZD50, (3, 2) = YTbbtoZD50, (3, 3) = ZTbbtoZD50})).(Matrix(3, 3, {(1, 1) = RX*Dr, (1, 2) = GX*Dg, (1, 3) = BX*Db, (2, 1) = RY*Dr, (2, 2) = GY*Dg, (2, 3) = BY*Db, (3, 1) = RZ*Dr, (3, 2) = GZ*Dg, (3, 3) = BZ*Db})).(Vector(3, {(1) = R_Tbb, (2) = G_Tbb, (3) = B_Tbb})) NULL

NULL

(Vector(3, {(1) = X_D50, (2) = Y_D50, (3) = Z_D50})) = (Matrix(3, 3, {(1, 1) = XTbbtoXD50, (1, 2) = YTbbtoXD50, (1, 3) = ZTbbtoXD50, (2, 1) = XTbbtoYD50, (2, 2) = YTbbtoYD50, (2, 3) = ZTbbtoYD50, (3, 1) = XTbbtoZD50, (3, 2) = YTbbtoZD50, (3, 3) = ZTbbtoZD50})).(Matrix(3, 3, {(1, 1) = RX, (1, 2) = GX, (1, 3) = BX, (2, 1) = RY, (2, 2) = GY, (2, 3) = BY, (3, 1) = RZ, (3, 2) = GZ, (3, 3) = BZ})).(Matrix(3, 3, {(1, 1) = Dr, (1, 2) = 0, (1, 3) = 0, (2, 1) = 0, (2, 2) = Dg, (2, 3) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = Db})).(Vector(3, {(1) = R_Tbb, (2) = G_Tbb, (3) = B_Tbb}))

NULL

(Vector(3, {(1) = X_D50, (2) = Y_D50, (3) = Z_D50})) = (Matrix(3, 3, {(1, 1) = RX_D50, (1, 2) = GX_D50, (1, 3) = BX_D50, (2, 1) = RY_D50, (2, 2) = GY_D50, (2, 3) = BY_D50, (3, 1) = RZ_D50, (3, 2) = GZ_D50, (3, 3) = BZ_D50})).(Matrix(3, 3, {(1, 1) = Dr, (1, 2) = 0, (1, 3) = 0, (2, 1) = 0, (2, 2) = Dg, (2, 3) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = Db})).(Vector(3, {(1) = R_Tbb, (2) = G_Tbb, (3) = B_Tbb}))

NULL

(Vector(3, {(1) = X_D50wht, (2) = Y_D50wht, (3) = Z_D50wht})) = (Matrix(3, 3, {(1, 1) = RX_D50, (1, 2) = GX_D50, (1, 3) = BX_D50, (2, 1) = RY_D50, (2, 2) = GY_D50, (2, 3) = BY_D50, (3, 1) = RZ_D50, (3, 2) = GZ_D50, (3, 3) = BZ_D50})).(Matrix(3, 3, {(1, 1) = Dr, (1, 2) = 0, (1, 3) = 0, (2, 1) = 0, (2, 2) = Dg, (2, 3) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = Db})).Camera_Neutral

NULL

Functions

   

NULL

Input Data

   

NULL

Solve for Camera to XYZ D50 and T

   

NULL


Download Camera_to_XYZ_Tcorr.mw

 

Hey I have a list of matrices and I want to define a proc that will search through the first matrices entries looking for zeroes, and if it finds one to move on to the next matrix in the list and look for zeroes and so on. if for some reason all matrices in the list have zeroes i would like the proc to answer with 0. otherwise I would like it to answer with the ndex of the first matrix wth all non-zero entries. I've played with things and occasionally made things work. But in general I do not have a solution. This is what I'e tried:

Things like this:

recu:=proc(y,n,q)
options trace;
local f,j,t,k;
t[y]:=y:
k:=y:
for f from 1 to n while y<=q do;
for j from 1 to n do;
if evalb(C[k](f,j)=0) then;
t[y+1]:=t[y]+1;
recu(t[y+1],n,q);
else next;
end if;
end do;
end do;
end proc;

like this:

recu:=proc(y,n,q)
local f,j,t,k;
global S;
options trace;
S:=0;
if y=q then return "no";
end if;
for f from 1 to n do;
for j from 1 to n do;
if evalb(C[y](f,j)=0) then;
recu(y+1,n,q);
S:=S+1;
end if;
end do;
end do;
end proc;

And I think I understand well why these are not workinging however I wondered what I can do that will be syntactically (sp?) correct.

Thanks

with(LinearAlgebra):
a:=Vector([1,2]);
b:=Matrix([[1,2],[1,2]]);

Say if I need a^T * b * a, I will do this:

VectorMatrixMultiply(Transpose(a), b);
VectorMatrixMultiply(%, a);

But this seems too long for such a simple matrix (and vector) computation. I am sure there must be an short way.

What if I need more computation, like

 

a^T * b * c*d*f*g* a, where c,d,f,g are other 2x2 matrices.
 If I were to use the above command, that'll take a long time to input.

Thanks,

Hi there

I have have a 18*18 matrix which almost each of its element are in symbolic form. Now I need to have all of its eigenvectors. Unfortunately when I use the "Eigenvalues()" function in maple i got nothing. In fact I got the error which comes below.

Error, (in content/polynom) general case of floats not handled

I need to know if there's a solution to eliminate the error? If not, what can I do to determine the eigenvectors and eigenvalues in symbolic form?

I'll be appreciated your help

Hi everyone,

I have a question regarding the derivation of tensors/matrices.
Let's assume for simplicity, that I have a vector (6x1) s and a matrix A (6x6)defining Transpose(s)*Inverse(A)*s. From this function I want to calculate the derivative w.r.t. s. My approach would be

restartwith(Physics):
with(LinearAlgebra):
Define(s,A)

Diff(
Transpose(s)*Inverse(A)*s, s)

As a result I get

though I'd rather expect something like Inverse(A)*s + Transpose(s)*Inverse(A)

Now as I'm pretty new to Maple, I can imagine that my approach is wrong, but I don't know any better and can't seem to get any information out of the help documents.

Thanks in advance for any of your suggestions!

 http://en.wikibooks.org/wiki/Linear_Algebra/

Representing_Linear_Maps_with_Matrices

 

how to calculate the first step

(2,0) -> (1,1,1) and (1,4) -> (1,2,0)

how to use maple command to get (1,1,1) and (1,2,0)

how to use maple command to calculate rep(h)

 

to get (0,-1/2,1) and (1,-1,0)

http://en.wikibooks.org/wiki/Linear_Algebra

/Representing_Linear_Maps_with_Matrices

I have two 6x1 Matrices which are the results of a calculation process in Maple. One with a set of equations and the other one with a set of variables: 

A := [0, f(x6), f(x6), 0, 0, f(x6)];

b := [x1, x2, x3, x4, x5, x6];

I'd like to solve the following system:

for i from 1 to 6 do

eq[i] := A[i] = b[i]:

od;

which is

eq[1] := 0 = x1;

eq[2] := f(x6) = x2;

eq[3] := f(x6) = x3;

...

 

If I type in the eqations manually, and execute "s := solve({eq[1],..,eq[6]},{x1,..,x6})" everything solves fine.

If I use the "for i from..." - structure, and execute "s := solve({eq[1],..,eq[6]},{x1,..,x6})" I get an empty space as solution.

I've tried to convert both matrices into lists, but it doesn't work.

Could it be that Maple doesnt know that x6 has to be the x6 in the function f(x6) ?

Can anyone tell me how to solve this please?

Hey Guys

I'm designing some stuff for controll theory

There I met following problem:

I want calculate the matrix A which is the product of X and G.

A is 6x6 matrix and all values in the diagonal  are ones, the rest is undefined and should be calculated

X is 6x6 diagonal matrix with the values X1...X6

and G is also 6x6 and have all these values.

 

now I'm stuck 'cause I haven't used Maple frequently (only some simple stuff up to now, nothing with matrices)

I'd really appreciate it, if someone can tell how to tackle this problem

Cheers

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