Linear transformations + writing of output matrix

[JamesGoh]

Homework Statement

Given the following defined transformation

$T(a + bt+ct^{2}) = (a+c) - (c+b)t + (a+b+c)t^{2}$

find the matrix with respect to the standard basis


From my understanding, the standard basis for a 3 element vector would
be

$(0,0,1)^{T} (0,1,0)^{T} (1,0,0)^{T}$

Homework Equations

T(u+v)= T(u) + T(v)

λT(v) = T(λv)

The Attempt at a Solution

okay, if I used the defined transformation, I get the following when I put any of the standard basis into the transformation

$T(0,0,1)^{T}=1 - t + t^{2} T(0,1,0)^{T} = 0 -t + t^{2} T(1,0,0)^{T} = 1 - 0 + t^{2}$

If I am correct, the matrix should be the following

1 -1 1
0 -1 1
1 0 1

However, the tutorial answers have it in the form

1 0 1

-1 -1 0

1 1 1

Shouldn't my answer be correct, since the $t$ and $t^{2}$ terms are different parts of a linear equation which is why they can't be in the same column ?

 

[rootX]

T(x) = A.x

where
x = [c b a]^T
A is the transformation matrix


The way you have put your A matrix, we would need to do following
T(p) = p.A
p = [c b a]

 

[Deveno]

evidently, the "standard basis" being referred to is {1,t,t²}.

that is: a+bt+ct²<---> (a,b,c)

(1,0,0) <--> 1
(0,1,0) <--> t
(0,0,1) <--> t²

with respect to a given basis, B = {v₁,...,v_n}, the matrix of a linear transformation T:V-->V, [T]_B, is the matrix such that:

[T]_B[v_j]_B = [T(v_j)]_B,

where the multiplication on the left is an nxn matrix multiplied by an nx1 matrix.

of course, [v_j]_B = [0,0,...1,...,0]_B, where the 1 is in the j-th place, which then gives us the j-th column of [T]_B.

 

[JamesGoh]

evidently, the "standard basis" being referred to is {1,t,t²}.

Does this case of "standard basis" only apply for any 3-element vector input written in the form of a parabolic equation ?

 

[HallsofIvy]

Yes. Your "standard basis", <1, 0, 0>, <0, 1, 0>, and <0, 0, 1>, would be correct for $R^3$. Of course, the assignment <1, 0, 0>->1, <0, 1, 0>->t, <0, 0, 1>->$t^2$ maps one to the other.