Linear transformations + writing of output matrix

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SUMMARY

The discussion centers on finding the matrix representation of the linear transformation T defined by T(a + bt + ct²) = (a + c) - (c + b)t + (a + b + c)t² with respect to the standard basis {1, t, t²}. The correct transformation matrix is derived from applying T to the standard basis vectors, resulting in the matrix:
1 0 1
-1 -1 0
1 1 1.
The confusion arises from the interpretation of the standard basis, which should be understood as {1, t, t²} rather than the conventional R³ basis.

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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}<br /> <br /> <br /> <br /> T(0,1,0)^{T} = 0 -t + t^{2}<br /> <br /> <br /> <br /> T(1,0,0)^{T} = 1 - 0 + t^{2}<br /> <br />

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 ?
 
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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]
 
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,...,vn}, the matrix of a linear transformation T:V-->V, [T]B, is the matrix such that:

[T]B[vj]B = [T(vj)]B,

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

of course, [vj]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.
 
Deveno said:
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 ?
 
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.
 

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