@Kitonum 

in fact, my application is for 

with(Groebner):
with(LinearAlgebra):
T := lexdeg([x1,x2,x3,x4,x5,x6,x7,x8],[e1,e2]);
GB := Basis([e1-x4*x2^3*x5/(x3*x1^4*x6), e2-x1*x8/(x3*x7)],T):

Error, (in LinearAlgebra:-Basis) invalid input: LinearAlgebra:-Basis expects its 1st argument, V, to be of type {Vector, {set(Vector), list(Vector)}} but received [e1-x4*x2^3*x5/(x3*x1^4*x6), e2-x1*x8/(x3*x7)]

however, got error, i have posted in another post, i do not understand why invalid input

@Kitonum 

in fact, there is no 

x1:=2:

x2:=3:

x3:=7:

x4:=17:

x5:=173:

x6:=709:

x7:=5347:

x8:=18713:

how to automatically distinct assign to x1,x2,..from only a,b,c,d?

@Kitonum 

i discover your method and sylvestorsolve (A,A,Y)

is the same

so what is the different between ker(A*X+X*A) = B and A*X+X*A = B ?

actually would like to solve for X in  ker(A*X+X*A) = B 

@Kitonum 

Y is also a known matrix, 

i make Y in terms of x because i want to find NullSpace^-1(Y) first and then subs values back into it.

is there a method to calculate  NullSpace^-1(Y) without making Y in terms of x

why Y must collinear with <1, 2, 3>, where do it come from? is it possible in general?

@Kitonum 

when i try it in another application

it return some values less than 1, 

never mind, i will think another method to deal with this.

may be the result matrix i derive that can not convert the original one with little difference

@Kitonum 

is it possible to solve when specfied some value, 

NewMatrix3 := Matrix([[x1,2,3], [2,3,0], [2,0,0]]):

simplify(solve({seq(seq((NewMatrix3 . EigenVector1(1..-1,i))[j]=(v[i]*EigenVector1(1..-1,i))[j], j=1..3), i=1..3)}, {seq(x||i, i=1..9)}));

@Kitonum 

why it evaluating a very long time

it has already run for 3 minutes, and it is still running

@Preben Alsholm

<Px, y> = <x, Py>

P is self joint

 is (H+H*)/2 a hermitian matrix when H is a non Hermitian matrix?

@Carl Love 

any book teach how to predict or approximate this little perturbation?

@Adri van der Meer 

the ev(4) is 3*3 matrix 

which are x, y, z?

actually the expected result of x, y, z should be in terms of variable test10

@Preben Alsholm 

use Eigenvalues, it has RootOf, why not in terms of variable test10

> New_EigenValue1 := Eigenvalues(MatrixMatrixMultiply(Transpose(NewMatrix3), NewMatrix3));
New_EigenValue1 := Vector(1, {(1) = RootOf(500000*_Z^7+(-500000*test10^2-402998250)*_Z^6+(-2503915066*test10+62844833450+221975850*test10^2)*_Z^5+(-3351902459500-15625257995*test10^2+179560038450*test10)*_Z^4+(391265707900*test10^2+82143407200000-4432825256000*test10)*_Z^3+(-1004104685500000+44430918230000*test10-3993649941000*test10^2)*_Z^2+(14116203710000*test10^2-154840724500000*test10+5938419770000000)*_Z-12500000*test10^2-13490160445000000+156500000*test10)})

then i try , got error too

> Matrix(NewMatrix3 -lambda*IdentityMatrix(7));
Error, (in Matrix) dimension parameters are required for this form of initializer

t:=1;
NewMatrix3 := Matrix([[test10, close3(t) , close3(t+1) , close3(t+2) , close3(t+3) , close3(t+4) , close3(t+5)],
[close3(t) , close3(t+1) , close3(t+2) , close3(t+3) , close3(t+4) , close3(t+5) ,0],
[close3(t+1) , close3(t+2) , close3(t+3) , close3(t+4) , close3(t+5) , 0,0],
[close3(t+2) , close3(t+3) , close3(t+4) , close3(t+5) , 0 , 0,0],
[close3(t+3) , close3(t+4) , close3(t+5) , 0 , 0 , 0,0],
[close3(t+4) , close3(t+5) , 0 , 0 , 0, 0,0],
[close3(t+5) , 0 , 0 , 0, 0, 0,0]]):
with(LinearAlgebra):
solve(Determinant(Matrix(NewMatrix3 – lambda*IdentityMatrix(7))), lambda);

@one_man 

how about this?

solve({3 = c1*1 + c2*2, 5 = c1*1 + c2*9, c1^2 + c2^2 = 1}, {c1,c2});

@Markiyan Hirnyk 

originally i do not want to show this system's real face, 

actually my system is

a1 := Diff(x1(s,t),s$2) = a*x1(s,t)+b*x2(s,t)+c*x3(s,t)+d*u(t);
a2 := Diff(x1(s,t),t)=x1(s,t);
b1 := Diff(x2(s,t),s$2) = e*x1(s,t)+f*x2(s,t)+g*x3(s,t)+h*u(t);
b2 := Diff(x2(s,t),t)=x2(s,t);
c1 := Diff(x3(s,t),s$2) = i*x1(s,t)+j*x2(s,t)+k*x3(s,t)+l*u(t);
c2 := Diff(x3(s,t),t)=x3(s,t);
sys := [a1, a2, b1, b2, c1, c2];
sol := pdsolve(sys);

@Carl Love 

plots:-spacecurve([Vector[row](1, {(1) = cos(60)*sin(t)-sin(60)*cos(t)}), Vector[row](1, {(1) = sin(60)*sin(t)+cos(60)*cos(t)}), Vector[row](1, {(1) = t})], t = 0 .. 6*Pi);

why blank diagram when draw this?

@Markiyan Hirnyk 

is it needed to change this add((X(tim[i])-x11[i])^2,i=1..N)+add((Y(tim[i])-y11[i])^2,i=1..N)+add((Z(tim[i])-z11[i])^2,i=1..N)+add((U(tim[i])-u11[i])^2,i=1..N)