What is the matrix for T relative to bases B and B' when T maps P3 to P4?

[robierob12]

Let T: P3 ----> P4
Im attempting to find the matrix for T relative to the bases B and B'

T maps P3 to P4

<br /> <br /> \begin{array}{l}<br /> T(ax^3 + bx^2 + cx + d) = (3x + 2)(ax^3 + bx^2 + cx + d) \\ <br /> B = \{ x^3 ,x^2 ,x,1\} \\ <br /> B^&#039; = \{ x^4 ,x^3 ,x^2 ,x,1\} \\ <br /> \end{array}<br /> <br /> <br /> <br />



Im able to do these porblems when say T:R3--->R3 and my bases are ordered vectors as a pair, triplets ect. but I am not seeing how to find the image of say X cubed under T. Do I just plug it in? then i get another polynomial, which I would need to write as a linear combination of the base B'?

If someone could show me how to find the image of the first one, T(x^3) I could go on from there.

 

[MathematicalPhysicist]

correct.
and just for the fun of it:
T(x^3)=(3x+2)x^3=3x^4+2x^3
T(x^2)=(3x+2)x^2=3x^3+2x^2
T(x)=(3x+2)x=3x^2+2x
T(1)=3x+2
which gives you somethin like this:

3000
2300
0230
0023
0002

 

[robierob12]

Thanks, for some reason I didnt see the where the substitutions went. I was trying to plug in my basis pieces into each x variable on the right hand side in the general polynomial.
Seems kinda funny now.

 

[robierob12]

correct.

which gives you somethin like this:

3000
2300
0230
0023
0002

just to make sure that I am seeing this correctly, the column entrys are also the scaler numbers in the linear combination (co efficients because I am working with a polynomial) of B'


So if base B' had a first vector of (2x^4) my matricie would look like this,

6000
2300
0230
0023
0002


Am I seeing this right?


3000
2300
0230
0023
0002


is my matrix for T relative to B and B'