I have two linear algebra texts [1, 2]  with examples of the process of constructing the transition matrix  that brings a matrix  to its Jordan form . In each, the authors make what seems to be arbitrary selections of basis vectors via processes that do not seem algorithmic. So recently, while looking at some other calculations in linear algebra, I decided to revisit these calculations in as orderly a way as possible.

First, I needed a matrix  with a prescribed Jordan form. Actually, I started with a Jordan form, and then constructed  via a similarity transform on . To avoid introducing fractions, I sought transition matrices  with determinant 1.

Let's begin with , obtained with Maple's JordanBlockMatrix command.

The eigenvalue  has algebraic multiplicity 6. There are sub-blocks of size 3×3, 2×2, and 1×1. Consequently, there will be three eigenvectors, supporting chains of generalized eigenvectors having total lengths 3, 2, and 1. Before delving further into structural theory, we next find a transition matrix  with which to fabricate .

The following code generates random 6×6 matrices of determinant 1, and with integer entries in the interval . For each, the matrix  is computed. From these candidates, one  is then chosen.

After several such trials, the matrix  was chosen as

for which the characteristic and minimal polynomials are

So, if we had started with just , we'd now know that the algebraic multiplicity of its one eigenvalue  is 6, and there is at least one 3×3 sub-block in the Jordan form. We would not know if the other sub-blocks were all 1×1, or a 1×1 and a 2×2, or another 3×3. Here is where some additional theory must be invoked.

The null spaces  of the matrices  are nested: , as depicted in Figure 1, where the vectors , are basis vectors.

The vectors  are eigenvectors, and form a basis for the eigenspace . The vectors , form a basis for the subspace , and the vectors , for a basis for the space , but the vectors  are not yet the generalized eigenvectors. The vector  must be replaced with a vector  that lies in  but is not in . Once such a vector is found, then  can be replaced with the generalized eigenvector , and  can be replaced with . The vectors  are then said to form a chain, with  being the eigenvector, and  and  being the generalized eigenvectors.

If we could carry out these steps, we'd be in the state depicted in Figure 2.

Next, basis vector  is to be replaced with , a vector in  but not in , and linearly independent of . If such a  is found, then  is replaced with the generalized eigenvector . The vectors  and  would form a second chain, with  as the eigenvector, and  as the generalized eigenvector.

Define the matrix  by the Maple calculation

and note

The dimension of  is 3, and of , 5. However, the basis vectors Maple has chosen for  do not include the exact basis vectors chosen for .

We now come to the crucial step, finding , a vector in  that is not in  (and consequently, not in  either). The examples in  are simple enough that the authors can "guess" at the vector to be taken as . What we will do is take an arbitrary vector in  and project it onto the 5-dimensional subspace , and take the orthogonal complement as .

A general vector in  is

A matrix that projects onto  is

The orthogonal complement of the projection of Z onto  is then . This vector can be simplified by choosing the parameters in Z appropriately. The result is taken as .

The other two members of this chain are then

A general vector in  is a linear combination of the five vectors that span the null space of , namely, the vectors in the list . We obtain this vector as

A vector in  that is not in  is the orthogonal complement of the projection of ZZ onto the space spanned by the eigenvectors spanning  and the vector . This projection matrix is

The orthogonal complement of ZZ, taken as , is then

Replace the vector  with , obtained as

The columns of the transition matrix  can be taken as the vectors , and the eigenvector . Hence,  is the matrix

Proof that this matrix  indeed sends  to its Jordan form consists in the calculation

The bases for , are not unique. The columns of the matrix  provide one set of basis vectors, but the columns of the transition matrix generated by Maple, shown below, provide another.

I've therefore added to my to-do list the investigation into Maple's algorithm for determining an appropriate set of basis vectors that will support the Jordan form of a matrix.

Download JordanForm_blog.mw

Need help for manipulating tensor with the physics package.

I ask some questions about this.  But each time, I am refer to the help pages.  If I ask again some help, it is because I can't not start with the information on the help file.  It is written for people that already know General Relativity (GR).

So this time, I have created a document (attach to this post) where I ask specific queations on manipulations.  My goal is to ccrreate a document that I will put on the Applications Center.  I promess that those who will help me on this will be cited in the document.  This way, I hope to create an introduction on how to use tensors for beginner like me.

Then, with this help, I am sure I will be able to better understand the help page of the packages.  I am doing this as someone who is starting to learn GR and have to be able to better understand the manipulations of tensor and getting the grasp of teh meaning of all those tensor.  For exemple, the concept of parallel transport on a curve surface.

Thank you in advance for all the troubling I give you with this demand.

Regards,Parallel_Transport.mw

--------------------------------------
Mario Lemelin

Maple 2015 Ubuntu 14.04 - 64 bits
Maple 2015 Win 7 -  64 bits
messagerie : mario.lemelin@cgocable.ca
téléphone :  (819) 376-0987

I using Maple 18.

I write below procedure,

Eqplot:=proc(a,b)
local y,x;

y:=a*x+b;

y,plot(y,x);

end proc:

when I use Eqplot(2,3) , then I will expect the result for both of equation 2x+3 and sketch y=2x+3, but unfortunately the result is only sketch y=2x+3.

please hint me about that how I can write a proc that result be both of 2x+3 and sketch 2x+3 in maple 18?

Hi,

I'm relatively new to using Maple.

I'm looking for information on how the "factor" function works. I printed its definition, and it refers to "factor/factor" and I can't find any more information on this. I'd like to know so that I could have more trust that it works correctly. Specifically, I'd like to be able to believe that if it does not factor a cubic, then the cubic is irreducible.

I'm specifically looking for rational roots of cubic polynomials. The "solve" function seems to work, and gives me the roots in terms of square roots, cubic roots and rationals. I have no idea why I should believe that "type(x,rational)" would work when the description of "x" is quite complicated.

Does anyone know anything about how "factor" works, or how "type" works when testing whether an expression evaluates to a rational number? Any information would be much appreciated.

Thanks,
Matt

How would you apply a permutation to a list?

So, with permutation [[1,3],[2,4]], given in disjoint cycles,

how would you apply it to [1,2,3,4] to obtain [3,4,1,2]?

Thanks

Perhaps the question is trivial, but I could not find the solution.

I am solving numerically an ODE (e.g., a simple harmonic oscillator) with the righthand side that contains a random part. For example it is

eq:=diff(a(t),t,t) + a(t) = (1+0.01*R(t))*cos(5*t);

where R(t) is a random function of t. How can I make such a function?

The naive attempt: 

eq:=diff(a(t),t,t) + a(t) = 3.*(1+0.1*rand()/1000000000000.)*cos(5.*t);

gave me a fixed (while random) value, e.g.

...

But, I need this coefficeint to be random each time step.

Any suggestions are very welcome!

after following a example , got error

                             2                  
               1   / d      \    1        2     2
               - m |--- x(t)|  - - m omega  x(t)
               2   \ dt     /    2              
Error, (in Mechanics:-LagrangeEqs) invalid input: subs received subst1, which is not valid for its 1st argument
Error, invalid input: Mechanics:-GeneralSol expects its 1st argument, eqs, to be of type list, but received eqs
Error, invalid input: rhs received sol, which is not valid for its 1st argument, expr
L;

Mechanics := module()
export SetVariables, LagrangeEqs, GeneralSol;
option package;
local subst1, subst2, varN, t;

SetVariables = proc( vars:: list, time )
local i;
t := time;
varN := nops( vars );
subst1 := {};
subst2 := {};
for i from 1 to var N do
subst1 := subst1 union
{vars[i](t) = q[i], diff(vars[i](t), t) = v[i]};
subst2 := subst2 union
{q[i] = vars[i](t), v[i] = diff(vars[i](t), t)};
end do;
print( subst1 );
print( subst2 );
NULL;
end proc;

LagrangeEqs := proc (L)
local i, l1, term1, term2;
l1 := subs(subst1, L):
for i to varN do
term1 := [seq(diff(subs(subst2, diff(l1, v[i])), t), i = 1..varN)]:
term2 := [seq(subs(subst2, diff(l1, q[i])), i = 1..varN)]:
end do;
[ seq(simplify(term1[i]-term2[i]) = 0, i = 1..varN) ];
end proc;

RayleighEqs := proc(L, R)
local i, l1, r1, term1, term2, term3;
l1 := subs( subst1, L ):
r1 := subs( subst1, R ):
for i from 1 to varN do
term1:=[seq(diff(subs(subst2, diff(l1, v[i])), t), i=1..varN)]:
term2:=[seq(subs(subst2, diff(l1, q[i])), i=1..varN)]:
term3:=[seq(subs(subst2, diff(r1, v[i])), i=1..varN)]:
end do:
[ seq(simplify(term1[i]-term2[i]+term3[i]), i=1..varN) ];
end proc;

LagrEqsII := proc( L, Q::list )
local i, l1, term1, term2;
l1 := subs(subst1, L):
for i to varN do
term1 := [seq(diff(subs(subst2, diff(l1, v[i])), t), i = 1 .. varN)]:
term2 := [seq(subs(subst2, diff(l1, q[i])), i = 1 .. varN)]:
end do;
[seq(simplify(term1[i]-term2[i]) = Q[i], i = 1 .. varN)];
end proc;

LagrEqsIII := proc (L, R, Q::list)
local i, l1, r1, term1, term2, term3;
l1 := subs(subst1, L):
r1 := subs(subst1, R):
for i to varN do
term1 := [seq(diff(subs(subst2, diff(l1, v[i])), t), i = 1 .. varN)]:
term2 := [seq(subs(subst2, diff(l1, q[i])), i = 1 .. varN)]:
term3 := [seq(subs(subst2, diff(r1, v[i])), i = 1 .. varN)]:
end do;
[seq(simplify(term1[i]-term2[i]+term3[i]) = Q[i], i = 1 .. varN)];
end proc;

GeneralSol := proc (eqs::list)
local i, initconds, eqs2;
initconds := NULL:
eqs2 := eqs[][]:
for i to varN do
initconds:=VarNames[i](0)=q[i], (D(VarNames[i]))(0)=v[i], initconds:
end do;
dsolve({initconds, eqs2});
end proc;

end module;

with(LibraryTools):
LibLocation := cat("c:\\Temp");
Save(Mechanics, LibLocation);
with(FileTools):
march('list',"c:\\Temp\\Mechanics.lib");
save(Mechanics, "c:\\Temp\\Mechanics.m");
read "c:\\Temp\\Mechanics.m";

with(Mechanics):
SetVariables([x], t);
L := (1/2)*m*diff(x(t), t)^2 - (1/2)*m*omega^2 * x(t)^2;
eqs := LagrangeEqs(L);
sol := GeneralSol( eqs );
X := unapply( rhs(sol), t );

I am doing a Calculus assignment and I can't find the commands for certain things.

1.Given the function f(x) = ((x+1)^2) / (1+x^2)

i) The domain of continuity of f(x)

ii) The intervals of increase and decrease of f(x) by using test points.

2. Use the IVT to prove existence of a root to the equation x^3 +10x^2 -100x +50=0 in the interval [-20,10]. Use again the IVT to show that there is a 1st root in [-17,-15], a 2nd toot in [0,1] and a 3rd root in [ 5,6]. Find or approximate those roots with Maple. (the bolded is what I need help).

I ran into a problem with the physics package that I subsequently solved. But I am wondering whether this would be a candidate for an SCR and/or be considered a bug.

The calculation I am trying is actually (so far) very simple.

I define a Hamiltonian H:

H := sqrt(p_^2*c^2+m^2*c^4); # note the square of vector p_

p_:=p1*_i+p2*_j+p3*_k;

H;

So far so good. Now I want to take the differentials of H against the components of p:

diff(H,p1) assuming c::real;

Hmm... I am not sure why the p2 and p3 still show up; but then, the product between the unit vectors should be 0 for different ones and one for equal unit vectors so maybe this is ok.

But H behaves weird: I can simplify it:

but if I try to do anything with it, it barfs:

dH+0;

Error, (in Physics:-Vectors:-+) invalid operation * between vectors _i and _j

As it turns out the issue is the square of the vector p_. Maple (or rather, Physics) does not recognize that it needs to expand p_^2 as p_.p_ and seems to treat is like p_*p_.

I would like Edgardo---& others more experienced with the Physics package than I am---to look at this. I do not understand the Physics package well enough to judge whether overloading the exponentiation operator to make this work is the right thing.

The example works once I replace p_^2 by p_.p_. But the ^2 notation is fairly standard usage so it feels slightly awkward.

Thanks,

Mac Dude.

Derivation_of_H.mw

The following illustrates a way to edit code when in Standard Worksheet.  Is there a bettery way?Thanks in advance,

Les

hello 

if you have a function lets say: 2x+1/4-x³

Now if you have to plot the graf, how should you know what the x and y shoud be? I mean you do like: plot(f(x),x=..

what should it be? i find it hard :(

Hello, forgive me if I used bad english, I am not a native speaker.

Anyhow: I have to decide the set of coordinates looking at the grafs intersections.  (I hope you understand that)

This is my function: f(x)=1/4*x³-x²-x+4

When you plot the function you see that the intersections is (-2,0) (2,0) and (4,0)

BUT I have to use a command to find these three intersections, plzz help me!

THANK YOU indeed.

[point of] intersection

Are there any commands in maple that will help me find a suitable function that approximates the numerical solution of:

Download PDEprob2_(2).mw

I am refering to the first graph, is there a way in maple to find an explicit suitable approximating function?

I.e, I want the function to have the same first graph obviously, it seems like addition of exponent and a line function, I tried plotting exp(-t)-0.3*t, it doesn't look like it approximates it very well. Any suggestion on how to implement this task in maple?

Thanks.

I'm have used a program to find the roots of a function 

f:=x*cos(x)-sin(x)*sin(x/1000);
/ 1 \
x cos(x) - sin(x) sin|---- x|
\1000 /

x_max:=50; x_min:=-50; step:=2; i_max:=(x_max-x_min)/step;
50
-50
2
50

j:=1:
for i from 0 to i_max by 1 do
x0:=x_min+i*step;
x[j]:=fsolve(f=0,x=x0);
j:=j+1;
end:

and my output was of the form of multiple "potential" roots and a bunch of which are the same. So I tried to get rid of the ones which were the same before actually finding the ones which ARE roots. To do that I done....

j := 1; for j to 50 do if x[j]-x[j+1] = 0 then ignore(x[j]) else print(x[j]) end if end do:

and it got rid of the ones which are of the above form but some roots are the same and seperated by more than 1 ... i.e x[ j ]= x[j + 2] or some other number. 

Basically I am trying to generalise the above for loop for all "numbers" instead of 1 but when I try some things the for loop doesnt like it. 

Any help would be good!

Say I have an equation of ax^2 + by^2 = 0,

I would like to plot a graph of y^2 against x^2...

How do i do that?