Finding f(T) without fixing a basis

Question

Suppose I'm given a linear operator $T(x_1,x_2)=(x_2,x_1+x_2)$. Let's call this representation A of the linear transformation. I can find a matrix for $T$ in the standard basis. It is $$\begin{bmatrix}0&1\\1&1\end{bmatrix}.$$ Given a polynomial, $f(x)=x^2+2x+3$, I can also find $f(T).$ It is $$\begin{bmatrix}4&3\\3&7\end{bmatrix}.$$ This is still in the standard basis however. How do I go from this matrix representation back to representation A.

My thoughts:

We have the following: $$f(T)(e_1)=4e_1+3e_2,$$ and $$f(T)(e_2)=3e_1+7e_2.$$ I'm really not sure where to go from here.

Answer 1

Note that by linearity $$ \begin{align} f(T)(x_{1},x_{2}) &=x_{1}f(T)(e_{1})+x_{2}f(T)(e_{2})\\ &=x_1(4e_{1}+3e_{2})+x_{2}(3e_{1}+7e_{2})\\ &=(4x_1+3x_2)e_{1}+(3x_1+7x_2)e_{2}\\ &=(4x_{1}+3x_{2}, 3x_{1}+7x_{2}). \end{align} $$ Alternatively one can note (perhaps more easily) that $$ f(T)(x_{1},x_{2})=\begin{bmatrix}4&3\\3&7\end{bmatrix} \begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix} $$

Answer 2

If you want $f(T)$ without fixing a base you must calculate

$f(T)=T^2+2T+3I$

where $T^2=T\circ T$