Linear operator find function

Question

Let $T(x,y)=(x,3x-y)$ be a linear operator on $R^2$ and let $f(x)=x^2-4$. Find $f(T)$

I'm very new to this and I'm stuck on what it's asking for. I'm guessing I make a matrix from the transformation and then use $rref$. I'm not sure where to go from there. Any guidance is appreciated.

I took matrix A and squared it and it returned the identity matrix. I then subtracted 4 times the identity matrix and got just (-3x, -3y). Can someone clarify what I'm doing incorrectly. Thanks

Answer 1

$T$ is realized by the matrix

$M =\begin{bmatrix} 1 & 0 \\ 3 & -1 \\ \end{bmatrix}$.

Then compute $M^2-4I$. Can you proceed ?

Answer 2

Alternatively, since $f(x)= x^2- 4$, $f(T)(\vec{x})= T^2(\vec{x})+ 4\vec{x}$ where $T^2$ is the composition of T with itself: $f(t)= T(T(\vec{x})+ 4\vec{x}$. $T(\vec{x})= T(\begin{bmatrix}x \\ y \end{bmatrix})= \begin{bmatrix}x \\ 3x- y\end{bmatrix}$ so $T^2(\vec{x})= \begin{bmatrix}x \\ 3x+ (3x- y)\end{bmatrix}= \begin{bmatrix}x \\ 6x- y\end{bmatrix}$.

So $f\left(T\left(\begin{bmatrix}x \\ y \end{bmatrix}\right)\right)= \begin{bmatrix}x \\ 6x- y\end{bmatrix}- \begin{bmatrix}4x \\ 4y\end{bmatrix}= \begin{bmatrix}-3x \\ 6x- 5y\end{bmatrix}$.