@Stephen Forrest 

how to change this expresssion in terms of numbers such as

permutation group

1 2 3

2 3 1

@Markiyan Hirnyk 

after tried evalc(Re(sys1)) = 0, it runs a long time,

i guess impossible, 

if i choose some good number for x and y to substitute, will it be possible?

any skill help to guess these number to make no imaginary part

sys1 := subs(x=?,subs(y=?, osys));
sys2 := subs(x=?,subs(y=?, osys));
sys3 := subs(x=?,subs(y=?, osys));

@Markiyan Hirnyk 

so, how to solve?

as my computer do not have maple in office, can the following code solve it?

res:=solve([Re(sys1)=0, Re(sys2)=0, Re(sys3)=0, Re(sys4)=0, Re(sys5)=0, Re(sys6)=0, Re(sys7)=0, Re(sys8)=0, Re(sys9)=0], {t1,t2,t3,t4,t5,t6,t7,t8,t9});

@Adri van der Meer 

this is the main point,
is it that two paramters need two equations in order to do inverse

how about

f := 2*x +  3*y

g := x^2 + y^3

if one equation is impossible, is it possible add extra arbitrary variable to make inverse function?

is this question be equivalent to that find back x and y in terms of f and g?

@Markiyan Hirnyk 

alpha:= (1/2)*(-y*t2-x*t1-y*t3+sqrt(y^2*t2^2+2*y*t2*x*t1+2*y^2*t2*t3+x^2*t1^2+2*x*t1*y*t3+y^2*t3^2-4*x*t4*y*t9-4*x*t4*y*t8-4*x^2*t4*t7-4*y^2*t5*t9-4*y^2*t5*t8-4*y*t5*x*t7-4*y^2*t6*t9-4*y^2*t6*t8-4*y*t6*x*t7))/(x*t4+y*t5+y*t6);
g := -y/x;
f := (-x+sqrt(x^2-x*y-2*y^2))/(2*y+x);
subs(p=f,subs(q=f,subs(x=p,subs(y=q,g))));
g := (1/2)*(-x+sqrt(x^2-4*y*x-4*y^2))/(x+y);
f := x*y;
gof := subs(p=f,subs(q=f,subs(x=p,subs(y=q,g))));
lhsgofoalpha := subs(q= alpha,subs(p=alpha, subs(x=p,subs(y=q,gof))));
foalpha := subs(p= alpha,subs(q=alpha,subs(x=p,subs(y=q,f))));
rhsgofoalpha := subs(x= foalpha,subs(y= foalpha, g));
osys := lhsgofoalpha = rhsgofoalpha;
sys1 := subs(x=3,subs(y=2, osys));
sys2 := subs(x=5,subs(y=1, osys));
sys3 := subs(x=1,subs(y=5, osys));
sys4 := subs(x=1,subs(y=2, osys));
sys5 := subs(x=2,subs(y=5, osys));
sys6 := subs(x=5,subs(y=2, osys));
sys7 := subs(x=2,subs(y=1, osys));
sys8 := subs(x=3,subs(y=5, osys));
sys9 := subs(x=5,subs(y=3, osys));
res:=solve([Re(sys1), Re(sys2), Re(sys3), Re(sys4), Re(sys5), Re(sys6), Re(sys7), Re(sys8), Re(sys9)], {t1,t2,t3,t4,t5,t6,t7,t8,t9});
eval(osys,res);
simplify(%);
`~`[lhs](select(evalb, res));

@Markiyan Hirnyk 

i understand now

however i can not see imaginary part in sys1

i guess that it occur in solve opeartion

i add Re(sys1)...., or using Re(t1)

return empty result,

how to do?

@Markiyan Hirnyk 

not 18 equations, i only use

sys1 := subs(x=3,subs(y=2, osys));
sys2 := subs(x=5,subs(y=1, osys));
sys3 := subs(x=1,subs(y=5, osys));
sys4 := subs(x=1,subs(y=2, osys));
sys5 := subs(x=2,subs(y=5, osys));
sys6 := subs(x=5,subs(y=2, osys));
sys7 := subs(x=2,subs(y=1, osys));
sys8 := subs(x=3,subs(y=5, osys));
sys9 := subs(x=5,subs(y=3, osys));

please use

alpha:= (1/2)*(-y*t2-x*t1-y*t3+sqrt(y^2*t2^2+2*y*t2*x*t1+2*y^2*t2*t3+x^2*t1^2+2*x*t1*y*t3+y^2*t3^2-4*x*t4*y*t9-4*x*t4*y*t8-4*x^2*t4*t7-4*y^2*t5*t9-4*y^2*t5*t8-4*y*t5*x*t7-4*y^2*t6*t9-4*y^2*t6*t8-4*y*t6*x*t7))/(x*t4+y*t5+y*t6);
g := -y/x;
f := (-x+sqrt(x^2-x*y-2*y^2))/(2*y+x);
subs(p=f,subs(q=f,subs(x=p,subs(y=q,g))));
g := (1/2)*(-x+sqrt(x^2-4*y*x-4*y^2))/(x+y);
f := x*y;
gof := subs(p=f,subs(q=f,subs(x=p,subs(y=q,g))));
lhsgofoalpha := subs(q= alpha,subs(p=alpha, subs(x=p,subs(y=q,gof))));
foalpha := subs(p= alpha,subs(q=alpha,subs(x=p,subs(y=q,f))));
rhsgofoalpha := subs(x= foalpha,subs(y= foalpha, g));
osys := lhsgofoalpha = rhsgofoalpha;
sys1 := subs(x=3,subs(y=2, osys));
sys2 := subs(x=5,subs(y=1, osys));
sys3 := subs(x=1,subs(y=5, osys));
sys4 := subs(x=1,subs(y=2, osys));
sys5 := subs(x=2,subs(y=5, osys));
sys6 := subs(x=5,subs(y=2, osys));
sys7 := subs(x=2,subs(y=1, osys));
sys8 := subs(x=3,subs(y=5, osys));
sys9 := subs(x=5,subs(y=3, osys));
res:=solve([sys1, sys2, sys3, sys4, sys5, sys6, sys7, sys8, sys9], real{t1,t2,t3,t4,t5,t6,t7,t8,t9});
eval(osys,res);
simplify(%);
`~`[lhs](select(evalb, res));

when i try it , it load a long time

@Markiyan Hirnyk 

i read it now.

most of them are the most primitive operation

it seems no polynomial or rational function can be associative

knowing x*y or x+y already enough

@Markiyan Hirnyk 

is it x*y and x+y the only associativity in the world?

@Preben Alsholm 

find simplify(%) compare by observation, i know t4 = 1

and deduce t1 = 1 from res

however, how to know t6 = 1 and t9 = 1?

@Carl Love 

how to make eigenvector be a matrix m*n with m < n?

would like to project to subspace which is a kind of eigenvectors

or project to basis of subspace

can eigenvectors be non-square?

@Axel Vogt 

what are the commands for computing a base for image(f)

@acer 

actually after get this list,

i will use degree to see which variables has negative power

[x^2, y^3, z^(-7)]

@one_man 

is it possible in general case?

assume the name variables may change too

but you can assume the format is x1, x1b

the most important step should be that extract 1/x2, 1/x3

or extract x2, x3 when they are division part

@Kitonum 

Moreover,

f := x1^2/((x2^3)*(x3^2));

would like to extract 1/x2 from 1/(x2^3) and 1/x3 from 1/(x3^2)

and then assign 1/x2 to x2b and assign 1/x3 to x3b

and then replace 1/(x2^3) with x2b^3 and 1/(x3^2) with x3b^2 in f

only in this example has x2 and x3, in real case, it is unknown how many variables all division part

how to automatically to do above for equations with division