Mathematics with Maple
S.Duzhin
Solutions for Lesson 8
> with(linalg):
Exercise 1.
a) Let e1 = 1, e2 = x-1/2, e3 = x^2 - x + 1/6 and D be the differentiation operator.
Then
D(e1)=0;
D(e2)=1=e1;
D(e3)=2*x-1=2*e2;
And the matrix of the differentiation operator D is as follows:
> A:=transpose(matrix([[0,0,0],[1,0,0],[0,2,0]]));
b) in a similar way, for the bases f1,f2,f3 we get
D(f1)= 0;
D(f2)=1=f1=[1,0,0];
D(f3)=x-1/2=f2+f1/2=[1/2,1,0];
> B:=transpose(matrix([[0,0,0],[1,0,0],[1/2,1,0]]));
we could do item b) in another way:
First, let us find the transition matrix:
f1=e1=[1,0,0];
f2=e2-e1/2=[-1/2,1,0];
f3=e3/2-e1/12=[-1/12,0,1/2];
we get the following matrix
> C:=transpose(matrix([[1,0,0],[-1/2,1,0],[-1/12,0,1/2]]));
and matrix B can be calculated as
> multiply(inverse(C),multiply(A,C));
Exercise 2.
> A:=matrix([[a,b],[c,d]]);
The bases in the vector space of all matrices 2x2 can be chosen, for example, as
> e1:=matrix([[1,0],[0,0]]);
> e2:=matrix([[0,1],[0,0]]);
> e3:=matrix([[0,0],[1,0]]);
> e4:=matrix([[0,0],[0,1]]);
And the action A takes these matrices into the matrices
A(e1)=[[a,0],[c,0]]=a*e1+c*e3;
A(e2)=[[0,a],[0,c]]=a*e2+c*e4;
A(e3)=[[b,0],[d,0]]=b*e1+d*e3;
A(e4)=[[0,0],[0,d]]=b*e3+d*e4;
> multiply(A,e1);
> multiply(A,e2);
> multiply(A,e3);
> multiply(A,e4);
in what we get the matrix B:
> B:=transpose(matrix([[a,0,c,0],[0,a,0,c],[b,0,d,0],[0,b,0,d]]));
Exercise 3.
Let calculate the images of vectors i,j,k under T:
> i:=vector([1,0,0]); j:=vector([0,1,0]); k:=vector([0,0,1]);
> crossprod(i,i);
> crossprod(i,j);
> crossprod(i,k);
Now we can find the matrix of operator T:
as
T(i)=0;
T(j)=k=[0,0,1];
T(k)=-j=[0,-1,0];
then the answer is
> A:=transpose(matrix([[0,0,0],[0,0,1],[0,-1,0]]));