Question

Linear Algebra:
Let T : P2 => P2 be the linear transform defined by: T(a+b*x+c*x^2) = b+2c*x

Find the matrix A of T with respect to the standard basis B = [1, x, x2].

Hint: You might use the fact that P2 is isomorphic to R3.

Answer 1

so it differentiates, thats fairly easy

Answer 2

I dont know how to do it though lol

Answer 3

i hate this stuff. u should b a fan of me if u can :)

Answer 4

you look at the way the function acts on the basic vectors

Answer 5

T(1) = 0 , which corresponds to a cordinate vector of (0,0,0)

Answer 6

T(1) = (0, 0, 0)
T(x) = (1, 0, 0)
T(x^2) = (0, 2, 0)

Answer 7

^yeh that

Answer 8

was that row reduced?

Answer 9

Then setup a matrix using the columns written in the standard basis.

Answer 10

then you have the first column of the matrix being the first vector

Answer 11

No, no row reduction in this problem.

Answer 12

and so on

Answer 13

\[\left[\begin{matrix}0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0\end{matrix}\right]\]

Answer 14

and that final result would be the matrix A of T?

Answer 15

The matrix A for the transformation T written with respect to the standard basis.

Answer 16

ahh ok very neat, just curious about one last thing. How did you form this:
T(1) = (0, 0, 0)
T(x) = (1, 0, 0)
T(x^2) = (0, 2, 0)
i know you're plugging in the bases of B into T but im not entirely sure whats happening inbetween

Answer 17

Plug each vector into the transformation and then write the resulting vector out with respect to the standard basis.

Answer 18

*Plug each basis vector

Answer 19

T(1) = 0*(1) + 0*(x) + 0*(x^2)
T(x) = 1*(1) + 0*(x) + 0*(x^2)
T(x^2) = 0*(1) + 2*(x) + 0*(x^2)

Answer 20

what i dont understand is how you got
000
100
020

Answer 21

Plug in the first basis element into the transformation. What do you get?

Answer 22

T(a+b*1+c*1^2) = b+2c*1?

Answer 23

Plug in the first basis element. The first basis element is 1.

Answer 24

i know i replaced x with 1

Answer 25

No, that's wrong. x is a place holder, these are vectors. You could just as well be dealing with a basis of apples and oranges.

Answer 26

This is the mapping [a, b, c] = [b, 2c, 0]

Answer 27

Mapping: [a, b, c] = [b, 2c, 0]
[1, 0, 0] = [0, 0, 0]
[0, 1, 0] = [1, 0, 0]
[0, 0, 1] = [0, 2, 0]

Answer 28

oh i see shouldnt it be [a, b, c] = [0, b, 2c]? though seeing a would go first or if a vector is empty does it go at the end of the mapped matrix

Answer 29

no, does anything get mapped to x^2?

Answer 30

ahhh i see now

Answer 31

ok i see how you arrived at the mapping for the beginning, but not this part
[1, 0, 0] = [0, 0, 0]
[0, 1, 0] = [1, 0, 0]
[0, 0, 1] = [0, 2, 0]
did you take the basis and put it with [a, b, c] = [b, 2c, 0] to arrive at the final result above?

Answer 32

A matrix is really just an ordered set of basis vectors that have been transformed and written with respect to the basis for the target space.

Answer 33

So perform the mapping for each basis vector and write it with respect to the standard basis.

Answer 34

i still dont get how you got
[1, 0, 0] = [0, 0, 0]
[0, 1, 0] = [1, 0, 0]
[0, 0, 1] = [0, 2, 0]

Answer 35

There is a bijection between any vector space and F^n for any field F. In this case we consider the field R. The point is you can perform a bijection between any vector in a given vector space and the coefficients in the linear combination.

Answer 36

1, becomes (1, 0, 0)
x, becomes (0, 1, 0)
x^2, becomes (0, 0, 1)

Answer 37

plug each standard basis element into the map
[a, b, c] = [b, 2c, 0]
[1, 0, 0] = [0, 0, 0]
[0, 1, 0] = [1, 0, 0]
[0, 0, 1] = [0, 2, 0]

Answer 38

ahh i see i see! thank you very much Alchemista again for your help today XD