Linear operators and matrices Let $B=\{(1,0),(1,1)\}$ be a basis of $\mathbb R^2$. Given the following matrix representation of an linear operator $T$ over the basis $B$:
$$[T]_B=
        \begin{pmatrix}
        -1 & 1 \\
        2 & 1  \\
        \end{pmatrix}
$$
How can we find the operator $T(x,y)$?
I don't know why my strategy doesn't work:
$T(x,y)=xT(1,0)+yT(0,1)=x(-1,2)+y(1,1)=(-x+y,2x+y)$.
I would like to know what's the standard method to find this operator.
Thanks in advance.
 A: Since $B$ isn't the standard basis then your strategy isn't correct. The standard way to find $T(x,y)$ is: let $P$ the change matrix from the canonical basis to the basis $B$ then
$$T(x,y)= P[T]_BP^{-1}(x,y)^T=(x+y,2x-y)$$
A: Your matrix means that:
$$T(1,0)=-1(1,0)+2(1,1)=(1,2)\quad\text{and}\quad T(1,1)=1(1,0)+1(1,1)=(2,1).$$
In this particularly easy case, we conclude that:
$$T(0,1)=T((1,1)-(1,0))=T(1,1)-T(1,0)=(2,1)-(1,2)=(1,-1).$$
Now, for all $(x,y)\in\mathbb{R}^2$,
$$T(x,y)=T(x(1,0)+y(0,1))=xT(1,0)+yT(0,1)=x(1,2)+y(1,-1)=(x+y,2x-y)$$
Don't confuse vectors of $\mathbb{R}^2$ and their coordinates in a basis.
Edit. To specifically answer your question in the comment:

So what is this matrix $[T]_B(1,0)=(−1,2)$?

Let me first reword it properly, since your product isn't exactly valid:

So what is this matrix $[T]_B\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}−1\\2\end{pmatrix}$?

The column $\begin{pmatrix}1\\0\end{pmatrix}$ can be understood as the coordinates in the basis $B$ of the vector of $\mathbb{R}^2$: $u=(1,0)$ (the first vector of the basis $B$). Then:
$$[T(u)]_B=[T]_B[u]_B=[T]_B\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}−1\\2\end{pmatrix}$$
is understood as _the coordinates of the vector $T(u)$ in the basis $B$ are $\begin{pmatrix}-1\\2\end{pmatrix}$, i.e.,
$$T(u)=-1(1,0)+2(1,1)=(1,2),$$
which should make sense.
If you asked:

So what is this matrix $[T]_B\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}1\\1\end{pmatrix}$?

I'd answer: The column $\begin{pmatrix}0\\1\end{pmatrix}$ can be understood as the coordinates in the basis $B$ of the vector of $\mathbb{R}^2$: $v=(1,1)$ (the second vector of the basis $B$). Then:
$$[T(v)]_B=[T]_B[v]_B=[T]_B\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}1\\1\end{pmatrix}$$
is understood as the coordinates of the vector $T(v)$ in the basis $B$ are $\begin{pmatrix}1\\1\end{pmatrix}$, i.e.,
$$T(v)=1(1,0)+1(1,1)=(2,1)$$
which should make sense too.
