It seems my book doesn\'t cover this sort of problem, any help would be apprecia

Question

It seems my book doesn't cover this sort of problem, any help would be appreciated.

Define V = P2, W = M2 times 2: V = (1 + x, x + x2 , 1 + x2), and Now define F : V rightarrow W by Then V and W are bases of V and W respectively, and F is a linear transformation. Find the matrix representation of F with respect to the standard bases of P2 and M2 times 2. Find the matrix representation of F with respect to (V, W). Let G : V rightarrow W be the linear transformation whose matrix representation with respect to (V. W) is: Find G(3 -x- 2x2).

Explanation / Answer

(a)

F(1+x) is the formula provided with a=1,b=1,c=0 so :

F(1+x)=[[ 3-5 , 1-0 ],[ 1+0, 1+1+0]] = [[-2,1],[1,2]] : coordinates (-2,1,1,2) in the standard base of M2x2

F(x+x^2) is the formlua provided with a=0,b=1,c=1 so :

F(x+x^2)=[[-5,-1],[2,2]] : coordinates (-5,-1,2,2) in the standard base of M2x2

F(1+x^2) is the formula provided with a=1,b=0,c=1 so :

F(1+x^2)=[[3,0],[1,2]] : coordinates (3,0,1,2) in the standard base of M2x2

The matrix is then this 3 vectors vertically :

[-2 -5 3]

[1 -1 0]

[1 2 1]

[2 2 2]

(b)

We have to express each F(1+x),F(x+x^2),F(1+x^2) = (-2,1,1,2),(-5,-1,2,2) and (3,0,1,2) in term of

(1,0,2,3),(0,-1,1,0),(0,1,0,2),(0,0,0,1) (elements of W written as coordinates)

(-2,1,1,2) = a(1,0,2,3) + b(0,-1,1,0) + c(0,1,0,2) + d(0,0,0,1)

a=-2

-b+c=1

2a+b=1 => b=1-2a=5 => c=1+b=6

3a+2c+d=2 => d=2-3a-2c=2+6-12=-4

So : (-2,1,1,2) = -2*(1,0,2,3)+5(0,-1,1,0)+6(0,1,0,2)-4(0,0,0,1) : coordinates (-2,5,6,-4) in W basis

(-5,-1,2,2) = a(1,0,2,3)+b(0,-1,1,0) + c(0,1,0,2) + d(0,0,0,1)

a=-5

-b+c=-1

2a+b=2 => b=2-2a=12 => c=-1+b=11

3a+2c+d=2 => d=2-3a-2c=2+15-22=-5

So : (-5,-1,2,2) = -5*(1,0,2,3)+12(0,-1,1,0)+11(0,1,0,2)-5(0,0,0,1) : coordinates (-5,12,11,-5) in W basis

(3,0,1,2) = a(1,0,2,3) + b(0,-1,1,0) + c(0,1,0,2) + d(0,0,0,1)

a=3

-b+c=0 => b=c

2a+b=1 => b=1-2a=-5 => c=-5

3a+2c+d=2 => d=2-3a-2c=2-9+10=3

So : (3,0,1,2) = 3*(1,0,2,3)-5(0,-1,1,0)-5(0,1,0,2)+3(0,0,0,1) : coordinates (3,-5,-5,3) in W basis

Finally the matrix of F with respect to (V,W) is :

[-2 -5 3]

[5 12 -5]

[6 11 -5]

[-4 -5 3]

(c)

We are given :

G(1+x)=(1,0,-1,0) (in W)

G(x+x^2)=(3,-1,2,1)

G(1+x^2)=(-2,1,0,-3)

So to find G(3-x-2x^2) we have to find a,b,c such that :

3-x-2x^2 = a(1+x)+b(x+x^2)+c(1+x^2) = a+c + x(a+b) + x^2(b+c)

a+c =3

a+b=-1

b+c=-2

So b-c=-4 and b+c=-2 => 2b=-6 => b=-3 and c=-2-b=1 and a=-1-b=2

So a=2,b=-3 and c=1 which means :

G(3-x-2x^2) = 2*G(1+x) -3 G(x+x^2) + G(1+x^2)

= 2*(1,0,-1,0)-3*(3,-1,2,1)+(-2,1,0,-3)

= (-9,4,-8,-6) (these are coordinates in W basis)

= -9*[[1 0],[2 3]] + 4[[0,-1],[1,0]] -8[[0,1],[0,2]] -6[[0,0],[0,1]](using elements of W)

=

[-9 -12]

[-14 -49]

There may be some typos, but the methodology is correct, ask if you have doubts.

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