Answer:
![[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]](https://tex.z-dn.net/?f=%5BT%5DEE%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%263%263%5C%5C6%264%264%5C%5C-2%263%264%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
First we start by finding the dimension of the matrix [T]EE
The dimension is : Dim (W) x Dim (V) = 3 x 3
Because the dimension of P2 is the number of vectors in any basis of P2 and that number is 3
Then, we are looking for a 3 x 3 matrix.
To find [T]EE we must transform the vectors of the basis E and then that result express it in terms of basis E using coordinates and putting them into columns. The order in which we transform the vectors of basis E is very important.
The first vector of basis E is e1(t) = 1
We calculate T[e1(t)] = T(1)
In the equation : 1 = a0
![[T(e1)]E=\left[\begin{array}{c}2&6&-2\\\end{array}\right]](https://tex.z-dn.net/?f=%5BT%28e1%29%5DE%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%266%26-2%5C%5C%5Cend%7Barray%7D%5Cright%5D)
And that is the first column of [T]EE
The second vector of basis E is e2(t) = t
We calculate T[e2(t)] = T(t)
in the equation : 1 = a1
![[T(e2)]E=\left[\begin{array}{c}3&4&3\\\end{array}\right]](https://tex.z-dn.net/?f=%5BT%28e2%29%5DE%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D3%264%263%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Finally, the third vector of basis E is 
![T[e3(t)]=T(t^{2})](https://tex.z-dn.net/?f=T%5Be3%28t%29%5D%3DT%28t%5E%7B2%7D%29)
in the equation : a2 = 1
Then
![[T(t^{2})]E=\left[\begin{array}{c}3&4&4\\\end{array}\right]](https://tex.z-dn.net/?f=%5BT%28t%5E%7B2%7D%29%5DE%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D3%264%264%5C%5C%5Cend%7Barray%7D%5Cright%5D)
And that is the third column of [T]EE
Let's write our matrix
T(X) = AX
Where T(X) is to apply the transformation T to a vector of P2,A is the matrix [T]EE and X is the vector of coordinates in basis E of a vector from P2
For example, if X is the vector of coordinates from e1(t) = 1
![X=\left[\begin{array}{c}1&0&0\\\end{array}\right]](https://tex.z-dn.net/?f=X%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D1%260%260%5C%5C%5Cend%7Barray%7D%5Cright%5D)
![AX=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]\left[\begin{array}{c}1&0&0\\\end{array}\right]=\left[\begin{array}{c}2&6&-2\\\end{array}\right]](https://tex.z-dn.net/?f=AX%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%263%263%5C%5C6%264%264%5C%5C-2%263%264%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D1%260%260%5C%5C%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%266%26-2%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Applying the coordinates 2,6 and -2 to the basis E we obtain
That was the original result of T[e1(t)]
