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]]));

[Maple Math]

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]]));

[Maple Math]

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]]));

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and matrix B can be calculated as

> multiply(inverse(C),multiply(A,C));

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Exercise 2.

> A:=matrix([[a,b],[c,d]]);

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The bases in the vector space of all matrices 2x2 can be chosen, for example, as

> e1:=matrix([[1,0],[0,0]]);

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> e2:=matrix([[0,1],[0,0]]);

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> e3:=matrix([[0,0],[1,0]]);

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> e4:=matrix([[0,0],[0,1]]);

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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);

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> multiply(A,e2);

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> multiply(A,e3);

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> multiply(A,e4);

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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]]));

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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]);

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> crossprod(i,i);

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> crossprod(i,j);

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> crossprod(i,k);

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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]]));

[Maple Math]