Linear Algebra - Linear Operator Suppose $\space T : \mathbb{R^2} \rightarrow \mathbb{R^2}$ is the linear operator defined by $T(x,y) = (x-y,x+y)$ $ \forall$  $(x,y)$ $\in \mathbb{R^2} $ then $T^{-1}(-1,-1) = (-1,0)$ 
Why does it equal $(-1,0)?$
What I did was $(-1-(-1),-1 + (-1))$ and got $(0,-2)$ then inversed it to get $(-2,0)$. What am I doing wrong? Any help is appreciated.
 A: Note that if $T^{-1}$ is an inverse, then
$$T^{-1}(T(x,y)) = (x,y) \implies T^{-1}(x-y, x+y) = (x,y).$$
If $u=x-y$ and $v=x+y$, then
$$T^{-1}(u,v) = \left(\frac{u+v}{2}, \frac{-u+v}{2}\right). $$
A: The issue is that you do not find the inverse by flipping the order of the coordinates. What you need to do is use the fact that $T^{-1}(x,y) = (a,b)$ if and only if $(x,y) = T(a,b)$.
Suppose that $T^{-1}(-1,-1) = (a,b)$, where we will find what $(a,b)$ must be. 
Since $T^{-1}(-1,-1) = (a,b)$, by applying $T$ to both sides:
$$
TT^{-1}(-1,-1) = T(a,b)
$$
we get that $(-1,-1) = T(a,b)$ since $TT^{-1}$ is the identity. 
We know that $T(a,b) = (a - b, a + b)$ so $(-1,-1) = (a - b,a+b)$ giving the system of linear equations
$$
\begin{align*}
-1 &= a - b\\
-1 &= a + b
\end{align*}
$$
Adding the equations gives that $-2 = 2a \Rightarrow a = -1$. Plugging back in we get $b = 0$. Thus, $(a,b) = (-1,0)$. That is, $T^{-1}(-1,-1) = (-1,0)$.
This can be generalized to any point $(x,y)$, not just $(-1,-1)$.
