LeeHoYeung

Mr. Ho Yeung Lee

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12 years, 234 days

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Seldom to ask question after retired math hobby Welcome August, February, July, May born girl And waited for her email to mavio@protonmail.com

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These are questions asked by LeeHoYeung

how to calculate basis <1,4,0>, <1,0,4> for eigenvalue 2;

how to calculate basis <1,0,1> for eigenvalue -1;

with(LinearAlgebra):
A := Matrix([[-2,1,1],[0,2,0],[-4,1,3]]);

sys1 := Eigenvalues(A)[1]*IdentityMatrix(3)-A;

sys1 := Eigenvalues(A)[2]*IdentityMatrix(3)-A;
sys1 := Eigenvalues(A)[3]*IdentityMatrix(3)-A;

 

B:=[<sys1[1,1],sys1[2,1],sys1[3,1]>,<sys1[1,2],sys1[2,2],sys1[3,2]>,<sys1[1,3],sys1[2,3],sys1[3,3]>,<0,0,0>];
LinearAlgebra:-Basis(B);

but not <1,4,0>, <1,0,4> for eigenvalue 2


 

invalid input: LinearAlgebra:-Basis expects its 1st argument, V, to be of type {Vector, set(Vector), list(Vector)

A:=<<5,5,5>|<1,2,3>|<-5,1,2>>;
Basis(A);
 

1.op(0,Expr) , op(1,Expr)

2. indets(eq1,{string,name})

3. type(varlist[ii], function)

read example
 
sph := <R*cos(u)*cos(v)|R*sin(u)*cos(v)|R*sin(v)>;
GK(sph); #Gauss Curvature
MK(sph); #Mean Curvature
 
how to find sph if slope is tan(u) ?
 
A. how to find xx1,xx2,xx3,yy1,yy2,yy3 that
Determinant(Matrix([[xx1,yy1,1],[xx2,yy2,1],[xx3,yy3,1]])) =(1/2)*aa*d*s*u+(1/2)*aa*d*s*a*t+(1/2)*d*v*u*t+(1/4)*d*v*a*t^2;
 
B. how to find x1,x2,x3,x4,y1,y2,y3,y4 that expand(
(x2 - x1)*(y4 - y3) - (y2 - y1)*(x4 - x3)) = (1/2)*d*s*aa*v+(1/2)*d*aa*v*u*t+(1/4)*d*aa*v*a*t^2+(1/2)*aa*d*s*u+(1/2)*aa*d*s*a*t+(1/2)*d*u^2*t+(3/4)*d*u*a*t^2+(1/4)*d*a^2*t^3;
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