Linear Transformation T: P2 to P3 & Matrix Representation

[jeffreylze]

Homework Statement

Let T: P₂ > P₃ denote the function defined by multiplication by x :T(p(x)) = xp(x). In other words, T(a+bx+cx²) = ax+bx²+cx³

(a) Show that T is a linear transformation.
(b) Find the matrix of T with respect to the standard bases {1,x,x²} for P₂ and {1,x,x²,x³} for P₃

Homework Equations

The Attempt at a Solution

I managed to prove that T is a linear transformation. With B, I have completely no idea how to go about, i checked my books but the examples given are not relevant. Please help.

 

[Dick]

Figure out where the standard basis vectors map. 1 in P2 is (1,0,0) in the given basis. That maps to x in P3, which is (0,1,0,0). Continue this for all of basis vectors in P2, scratch your head and figure out how to write a matrix which does the same thing. THINK about it.

 

[jeffreylze]

1 in P2 is (1,0,0) in the given basis. That maps to x in P3, which is (0,1,0,0).

I don't really understand this part, how does 1 in p2 maps to x in p3 ?

 

[Dick]

T(p(x))=x*p(x). So if p(x)=1, then T(1)=x*1=x. 1 in P2 maps to x in P3.

 

[jeffreylze]

Oh, so

p(x) = x , x in p₂ will map it to x² in p₃? - (0,0,1,0)
p(x) = x², X² in p₂ will map it to x³ in p₃ - (0,0,0,1)

So that gives me

A = [0 0 0;
1 0 0
0 1 0;
0 0 1] ?

But is there a faster way to do this? I came across this equation while looking for extra info online, [T(u)]_C=A_B . Will that simplify the method?

 

[Dick]

That's exactly right. What was "not faster" about the way you already did it? Once you understood the problem you solved it in two minutes. Why complicate it?

 

[jeffreylze]

Using that method, i tried solving this question but to no avail :

Find the matrix representation of T:P₁ > P₂ with respect to bases B = {1,x} and C {1,x,x²} where

T(p) = (x+2)p for p\inP₁

p = a₀ + a₁x

T(1) = (x+2)
T(x) = (x²+2x)

and I don't know how to map that to P₂

 

[Dick]

1 in P2 is (1,0). (x+2) in P2 is (2,1,0). PLEASE say you knew that. What is x in P1 and what is (x^2+2x) in P2? Now write down a 2x3 matrix and start filling in the columns. Your best tool is actually thinking about the problem. There is no magic formula.

 

[jeffreylze]

I am such a douche + a slow learner =/ Yeah, i get it (x+2) in P2 is (2,1,0) (it is just the same thing like the previous example! careless me) x in p1 is just (0,1) so (x^2+2x) in P2 will be (0,2,1). So the matrix will be A = (2,1,0 ; 0,2,1)

So a different basis will still be the same. Say B = {1,x-2} and C = {2,x,x^2} for P1 and P2 respectively.

T(1) = (x+2)
T(x-2)= (x^2-4)

1 in P1 will be (1,0) , and will map to (1,1,0) in P2

(x-2) in P2 will be (0,1) and will map to (-2,0,1) in P2

Yeap, eureka, i think i got it. Also, do you think these steps will be sufficient to answer exams questions? Or do I need a more rigid method/calculations ?

 

[Dick]

You are doing exactly what you are supposed to do. Compute the product, figure out the components in the given bases and deduce the matrix.