Answer:
[tex][T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right][/tex]
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
[tex]T(1)=(2.1+3.0+3.0)+(6.1+4.0+4.0)t+(-2.1+3.0+4.0)t^{2}=2+6t-2t^{2}[/tex]
[tex][T(e1)]E=\left[\begin{array}{c}2&6&-2\\\end{array}\right][/tex]
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
[tex]T(t)=(2.0+3.1+3.0)+(6.0+4.1+4.0)t+(-2.0+3.1+4.0)t^{2}=3+4t+3t^{2}[/tex]
[tex][T(e2)]E=\left[\begin{array}{c}3&4&3\\\end{array}\right][/tex]
Finally, the third vector of basis E is [tex]e3(t)=t^{2}[/tex]
[tex]T[e3(t)]=T(t^{2})[/tex]
in the equation : a2 = 1
[tex]T(t^{2})=(2.0+3.0+3.1)+(6.0+4.0+4.1)t+(-2.0+3.0+4.1)t^{2}=3+4t+4t^{2}[/tex]
Then
[tex][T(t^{2})]E=\left[\begin{array}{c}3&4&4\\\end{array}\right][/tex]
And that is the third column of [T]EE
Let's write our matrix
[tex][T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right][/tex]
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
[tex]X=\left[\begin{array}{c}1&0&0\\\end{array}\right][/tex]
[tex]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][/tex]
Applying the coordinates 2,6 and -2 to the basis E we obtain
[tex]2+6t-2t^{2}[/tex]
That was the original result of T[e1(t)]
