# Linear Trans. & Bases

1. May 3, 2007

### 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

$$\begin{array}{l} T(ax^3 + bx^2 + cx + d) = (3x + 2)(ax^3 + bx^2 + cx + d) \\ B = \{ x^3 ,x^2 ,x,1\} \\ B^' = \{ x^4 ,x^3 ,x^2 ,x,1\} \\ \end{array}$$

Im able to do these porblems when say T:R3--->R3 and my bases are ordered vectors as a pair, triplets ect. but Im 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.

2. May 3, 2007

### 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:
Code (Text):

3000
2300
0230
0023
0002

3. May 3, 2007

### 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.

4. May 6, 2007

### robierob12

just to make sure that im seeing this correctly, the column entrys are also the scaler numbers in the linear combination (co efficients because im 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'

Last edited: May 6, 2007