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I’m trying to figure out how to find a basis for a subspace, V, of Rdefined by V = {(x, y, z)l(2x-3y+6z=0)}

 

I’m using the student linear algebra module for maple 17

 

I’ve tried defining the subspace and asking for the basis of V but I always get an error code.

 

I’ve tried consulting the maple website and looking through their help menu, but can’t find anything that answers how to find a basis... At least a basis from the subspace defined in my problem.

I know how to find a basis for the subspace by hand but not with maple.

Any help will be greatly appreciated. 

Hi all 

I have the following segment of maple program which belongs to time delay systems dynamic. here C=X-X0-G.Z-X.Dtau.P+X.Dtau.Z-U.P, is a matrix(vector) which comes from reordering the system terms and my goal is to minimizing J:=X.E.Transpose(X)+U.E.Transpose(U), subject to constraint C=0, but i don't know how to do so.

I will be so grateful if anyone can guide me

best wishes

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department


restart:
with(Optimization):
with(LinearAlgebra):
macro(LA= LinearAlgebra):
L:=1:  r:=2:  tau:= 1:
interface(rtablesize= 2*r+1):

Z:= Matrix(
     2*r+1, 2*r+1,
     [tau,
      seq(evalf((L/(2*(iz-1)*Pi))*sin(2*(iz-1)*Pi*tau/L)), iz= 2..r+1),
      seq(evalf((L/(2*(iz-1-r)*Pi))*(1-cos(2*(iz-1-r)*Pi*tau/L))), iz= r+2..2*r+1)
      ],
     scan= columns,
     datatype= float[8]
);
                        
Dtau00:= < 1 >:
Dtau01:= Vector[row](r):
Dtau02:= Vector[row](r):
Dtau10:= Vector(r):
Dtau20:= Vector(r):

Dtau1:= LA:-DiagonalMatrix([seq(evalf(cos(2*i*Pi*tau/L)), i= 1..r)]):
Dtau2:= LA:-DiagonalMatrix([seq(evalf(sin(2*i*Pi*tau/L)), i= 1..r)]):
Dtau3:= -Dtau2:
Dtau4:= copy(Dtau1):

Dtau:= < < Dtau00 | Dtau01 | Dtau02 >,
         < Dtau10 | Dtau1  | Dtau2  >,
         < Dtau20 | Dtau3  | Dtau4  > >;
 
P00:= < L/2 >:
P01:= Vector[row](r):
P02:= Vector[row](r, j-> evalf(-L/j/Pi), datatype= float[8]):
P10:= Vector(r):
P20:= Vector(r, i-> evalf(L/2/i/Pi)):
P1:= Matrix(r,r):
P2:= LA:-DiagonalMatrix(P20):
P3:= LA:-DiagonalMatrix(-P20):
P4:= Matrix(r,r):

P:= < < P00 | P01 | P02 >,
      < P10 | P1  | P2  >,
      < P20 | P3  | P4  > >;

interface(rtablesize=2*r+1):    # optionally
J:=Vector([L, L/2 $ 2*r]):      # Matrix([[...]]) would also work here

E:=DiagonalMatrix(J);

X:=  Vector[row](2*r+1,symbol=a);
U:=Vector[row](2*r+1,symbol=b);

X0:= Vector[row](2*r+1,[1]);
G:=Vector[row](2*r+1,[1]);
C:=simplify(X-X0-G.Z-X.Dtau.P+X.Dtau.Z-U.P);

Z := Matrix(5, 5, {(1, 1) = 1., (1, 2) = 0., (1, 3) = 0., (1, 4) = 0., (1, 5) = 0., (2, 1) = 0., (2, 2) = 0., (2, 3) = 0., (2, 4) = 0., (2, 5) = 0., (3, 1) = 0., (3, 2) = 0., (3, 3) = 0., (3, 4) = 0., (3, 5) = 0., (4, 1) = 0., (4, 2) = 0., (4, 3) = 0., (4, 4) = 0., (4, 5) = 0., (5, 1) = 0., (5, 2) = 0., (5, 3) = 0., (5, 4) = 0., (5, 5) = 0.})

Dtau := Matrix(5, 5, {(1, 1) = 1, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (1, 5) = 0, (2, 1) = 0, (2, 2) = 1., (2, 3) = 0, (2, 4) = 0., (2, 5) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 1., (3, 4) = 0, (3, 5) = 0., (4, 1) = 0, (4, 2) = -0., (4, 3) = -0., (4, 4) = 1., (4, 5) = 0, (5, 1) = 0, (5, 2) = -0., (5, 3) = -0., (5, 4) = 0, (5, 5) = 1.})

P := Matrix(5, 5, {(1, 1) = 1/2, (1, 2) = 0, (1, 3) = 0, (1, 4) = -.318309886100000, (1, 5) = -.159154943000000, (2, 1) = 0, (2, 2) = 0, (2, 3) = 0, (2, 4) = .1591549430, (2, 5) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 0, (3, 4) = 0, (3, 5) = 0.7957747152e-1, (4, 1) = .1591549430, (4, 2) = -.159154943000000, (4, 3) = 0, (4, 4) = 0, (4, 5) = 0, (5, 1) = 0.7957747152e-1, (5, 2) = 0, (5, 3) = -0.795774715200000e-1, (5, 4) = 0, (5, 5) = 0})

E := Matrix(5, 5, {(1, 1) = 1, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (1, 5) = 0, (2, 1) = 0, (2, 2) = 1/2, (2, 3) = 0, (2, 4) = 0, (2, 5) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 1/2, (3, 4) = 0, (3, 5) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 0, (4, 4) = 1/2, (4, 5) = 0, (5, 1) = 0, (5, 2) = 0, (5, 3) = 0, (5, 4) = 0, (5, 5) = 1/2})

X := Vector[row](5, {(1) = a[1], (2) = a[2], (3) = a[3], (4) = a[4], (5) = a[5]})

U := Vector[row](5, {(1) = b[1], (2) = b[2], (3) = b[3], (4) = b[4], (5) = b[5]})

X0 := Vector[row](5, {(1) = 1, (2) = 0, (3) = 0, (4) = 0, (5) = 0})

G := Vector[row](5, {(1) = 1, (2) = 0, (3) = 0, (4) = 0, (5) = 0})

C := Vector[row](5, {(1) = 1.500000000*a[1]-2.-.1591549430*a[4]-0.7957747152e-1*a[5]-.5000000000*b[1]-.1591549430*b[4]-0.7957747152e-1*b[5], (2) = a[2]+.1591549430*a[4]+.1591549430*b[4], (3) = a[3]+0.7957747152e-1*a[5]+0.7957747152e-1*b[5], (4) = a[4]+.3183098861*a[1]-.1591549430*a[2]+.3183098861*b[1]-.1591549430*b[2], (5) = a[5]+.1591549430*a[1]-0.7957747152e-1*a[3]+.1591549430*b[1]-0.7957747152e-1*b[3]})

(1)

J:=X.E.Transpose(X)+U.E.Transpose(U);

J := a[1]^2+(1/2)*(a[2]^2)+(1/2)*(a[3]^2)+(1/2)*(a[4]^2)+(1/2)*(a[5]^2)+b[1]^2+(1/2)*(b[2]^2)+(1/2)*(b[3]^2)+(1/2)*(b[4]^2)+(1/2)*(b[5]^2)

(2)

Minimize(J,{C=0});






Error, (in Optimization:-NLPSolve) invalid arguments

 

#XP:=-.015+X[1]+add(X[l+1]*f1(l)+X[r+l+1]*f2(l), l= 1..r):
#plot([XP,T1], t= 0..1);#,legend= "Solution Of x(t) with r=50"):

 

 

 

 

 

 

Download work1.mwswork1.mws

Hi all

In matlab software we have a command namely fmincon which minimizes any linear/nonlinear algebric equations subject to linear/nonlinear constraints.

Now my question is that: what is the same command in maple?or how can we minimize linear/nonlinear function subject to linear/nonlinear constraints in maple?

thanks a lot

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department

The result of the following lines in Maple is V := [5., 3.,5. 11., 6.] that I think should be V := [5., 3.,5. 0., 6.], is there somthing wrong?

 

v1:= Vector([-28., -63., -17., -55., 17.], datatype= float[8]):

V:= LinearAlgebra:-Modular:-Mod(11, v1, float[8]);

bug.mw



Hi,

I'm currently writing a programme for synchronising automatas, its creates an array and adds words or matrices to the array that aren't already in. I am currently getting an error when i try and run my procedure though and i'm unsure of the problem, any help would be appreciated, here is my code so far.

Thanks!

proc_cerny1:=proc(A::Matrix,B::Matrix,C::Vector[row],N)
local x, S, i, j, T, R, y, found;
x:=2^N;
S:=Array(0..(x-1));
S[0]:=C;
i:=0;
j:=1;
found:=false;

while (i<(x-1) and i<>j) do
T:=S[i].A;
for y from 0 to j do
if LinearAlgebra:-Equal(S[y],T) then
found:=true;
if (found=false) then
S[j]:=T;
j:=j+1;
end if:
end if:
od:

R:=S[i].B;
for y from 0 to j do
if LinearAlgebra:-Equal(S[y],T) then
found:=true;
if (found=false) then
S[j]:=R;
j:=j+1;
i:=i+1;

end if:
end if:
od:
od:
print(S);
end proc:

 

The error i'm getting is when i input this:

proc_cerny1(Matrix([[1,0,0],[1,0,0],[1,0,0]]),Matrix([[0,1,0],[0,0,1],[0,0,1]]),Vector(1..3,1,orientation=row),3);

and the error is:

Error, (in LinearAlgebra:-Equal) invalid input: LinearAlgebra:-Equal expects its 1st argument, X, to be of type {Matrix, Vector} but received 0

 

I have to make a plot for experimental data where x corresponds to time and y is the concentration. So I loaded the values in two lists, ran fit to least squares and got the best fitting curve (it's linear). However, the line has to start at (0,0). How can I Maple to do that for me?

Hello,
my question may be simple but I don't find the answer in any help guide.
when I define a function I cannot use a linearalgebra expression such as Trace.
Here is an example of what I would like to do:




If anyone can help me...
Thank you

I have a statistics task that is very complicated to me, and I would really appretiate some help!

I believe that I have completed task a, but task b and c seems to be too hard for me to grasp.

Here is a presentation of the task text:

 

xi 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
yi 12.3 12.8 16.8 17.9 20.5 22.7 24.9 27.5 26.9 29.4 30.3 34.2 36.4 36.7 41.1

Assume that the data represents independent realizations of a "linearlnormal" model.

Yi = α + β xi + ei i = 1..15,

where ei˜N(0,σ), where sigma is unknown.

 

(a) create a scatterplot of the data {f(xi,yi)}15 ; i=1
Then calculate the empiric correlation coefficient.

(b) Find the minimum least square estimate for α & β.

(c) Create a 95% confidence interval for the parameter β. What does this confidence interval express?

Is there anyone out there who are able to help me with some maplecode?


thanks in advance,

Soeta

Hi,

I have a Matrix whose entries are polynomials in several formal parameters (the matrix is sparse and the polynomials are rather simple, though inverses of the parameters may also arise).
Then, when I compute the kernel with LinearAlgebra-NullSpace, maple naturally gives a basis of solutions over the same ring of polynomials.

Now for some reason there are some parameters that I don't want to see in the solutions (all but two of them, actually).

How can I compute the part of the kernel that lives in $\mathbb{Z}[a,a^{-1}, b,b^{-1}]$, i.e. that involves only the first two parameters?

Thank you,

NoThik

Edit : the coefficients of the polynomials are integers, and I expect the kernel elements to have integer coefficients as well.

I have a set of around 60 linear equations with symbolic coefficients. ie

 

a*x1 + b*x2 + ... + c*x60 = 1

x1 + (c-a)*x2 + ... + d*x60 = 0

...

c*x1 + d*x2 + ... + b*x60 = a

 

The coefficients a,b,c,d are functions of x1...x60. I am trying to find the values of these coefficients. When I had a smaller set of equations I was solving them symbolically to find x1...x60 in terms of a,b,c,d and then using this solution to solve for a,b,c,d. I can no longer solve the set of equations symbolically as it is too large. How do I find the coefficients? I had some sort of optimization routine in mind.

How would I write a function that produces true if a vector/ line satisfies 3 linear inequalities and false if it does not satisfy all 3 linear inequalities? Ie
If the vector <m,n> satisfies ax+by+c>0, dx+ey+f>0 and gx+hy+i>0 then the function returns true, and if it does not satisfy one or more then it returns false? Thanks very much for your time.

I have total nine equations,out of which some are linear and some of them are nonlinear.I am a new user of maple.The nine equations are

c2+vc = .34,

c2+c3 = .519,

t8+c3-c4-vc = .132,

t1+t5-c2+0.0435=0,

t1-t2-c2+c3-1.334*vc = 0,

t2+t6-c3-c4+0.0435+1.334*vc=0,

-112.5*t1+112.5*t5+300*c2^2+1000*vc = 38.25,

112.5*t1-112.5*t2-300*c2^2+300*c3-300*c3^2-2000*vc+23.4 = 0,

-112.8*t8+112.5*t2-112.5*t6+300*c4^2-300*c3+300*c3^2+1000*vc+14.85 = 0

Hi all,

I need to solve this linear system:

 

R11+R12+fx=1

R21+R22+fy=1

R31+R32+fz=1

R11+R21+R31=1

R12+R22+R32=1

R11*R22*fz+R12*fy*R31+fx*R21*R32-R31*R22*fx-R32*fy*R11-fz*R21*R12=1

 

It is a solution for a 3D rotation matrix in which I know the last coloumn (fx,fy,fz) and I need to find the other six elements of the matrix R11,R12...Obvioulsy there are infinite solution, but in particular...

It's interesting that every continuous piecewise linear function can be specified by one explicit equation with absolute values​​. The procedure JoggedLine carries out such conversion.

Formal arguments of the procedure: 

A - a list of the coordinates of the vertices of the polyline or the continuous piecewise linear expression defined on the entire real axis.

B (optional) - a point on the left "tail"...

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