Finding Linear Transformations in Polynomial Vector Spaces

[dim&dimmer]

[/b]1. Homework Statement [/b]
Let P_2 be the set of all real polynomials of degree no greater than 2.
Show that both B:={1, t, t^2} and B':= {1, 1-t, 1-t-t^2} are bases for P_2.

If we regard a polynomial p as defining a function R --> R, x |--> p(x), then p is differentiable, and
D: P_2 --> P_2, p |--> p' = dp/dx - defines a linear transformation.
Find the matrix of D with respect to the bases
(i) B in both the domain and co-domain
(ii) B in the domain and B' in the codomain.
(iii) B' in the domain and B in the codomain
(iv) B' in both the domain and codomain.

[/b]2. The attempt at a solution[/b]
I used 3x3 matrices and row reduction for B' to show that they are both bases, i.e. 3 pivot variables in reduced row form. B was just the identity matrix.
Its the second part that I don't understand. For me, the domains of B and B' are all real numbers, or have I already found the matrices of D in both domains by showing with row echelon matrices that B and B' are bases.
This is very confusing and my mind is now doing many laps around the same circuit.
Any clarification would be great.

 

[HallsofIvy]

Suppose T is a Linear transformation from vector space U with basis B_u to vector space V with basis B_v. To find the matrix representing T using basis B_u in the domain and basis B_v in the codomain, do the following:

1) Apply T to the first vector in basis B_u. The result, of course, will be in V and can be written as a linear combination of the vectors in basis B_v. The coefficients in that linear combination are the numbers in the first column of the matrix.

2) Apply T to the second vector in basis B_u. The result, of course, will be in V and can be written as a linear combination of the vectors in basis B_v. The coefficients in that linear combination are the numbers in the second column of the matrix.

Continue applying T to each basis vector in turn to get each column of the matrix.

For example, the first vector in B is "1". Differentiating that gives 0. That, written in the B basis is (0)1+ (0)x+ (0)x². The first column consists entirely of 0's. The second vector in B is "x". Differentiating that gives 1. That, written in the B basis is (1)(1)+ (0)x+ (0)x². The second column is <1 0 0>. The third vector in B is "x²". Differentiating that gives 2x. That, written in the B basis is (0)1+ (2)x+ (0)x². The third column is <0 2 0>. (i) is
[tex]\left[\begin{array}{ccc}0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0\end{array}\right][/tex]

Notice that matrix has determinant 0 and is not invertible. That is because differentiation is not one-to-one and so is not invertible.

 

[dim&dimmer]

so...
B matrix is [tex]
\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]
[/tex]
is the domain?
codomain of B is
[tex]
\left[\begin{array}{ccc}0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0\end{array}\right]
[/tex]
as you said
B' domain is
[tex]
\left[\begin{array}{ccc}1 & 1 & 1 \\ 0 & -1 & -1 \\ 0 & 0 & -1\end{array}\right]
[/tex]
codomain of B'
[tex]
\left[\begin{array}{ccc}0 & -1 & -1 \\ 0 & 0 & -2 \\ 0 & 0 & 0\end{array}\right]
[/tex]
Are these ok??

To find the matrices D with respect to the 4 part question, is it just a matter of solving the augmented matrices formed in each part, so that part i) would be the codomain matrix of B and part ii)the codomain matrix of B', given the identity natrix is the domain of B?

Why this is scary is because of the magnitude of how wrong I could be lol.