Deducing formula for a linear transformation The question I'm answering is as follows:
Let $ T: \mathbb{R}^2  \rightarrow \mathbb{R}^2$ be a linear transformation such that $ T(1,1) = (2,1) $ and $ T(0,2) = (2,8) $. Find a formula for $ T(a,b) $ where $ (a,b) \in \mathbb{R}^2 $.
Earlier we proved that $\{(1,1), (0,2)\}$ spans $\mathbb{R}^2$. I used this when trying to find a formula for $T$. My working is:
$T(a(1,1) + b(0,2)) = aT(1,1) + bT(0,2) $ 
Because $T$ is linear. Thus:
$ T(a(1,1) + b(0,2)) = a(2,1) + b(2,8) = T(a,b)$
Is this correct? It seems a bit too easy and so I'm wondering if I missed anything.

The matrix representation of $T$ with respect to the bases $\{(1,1),(0,2)\}$ (for the domain) and $\{(1,0),(0,1)\}$ (for the co-domain) is
$$\left( \begin{array}{cc}
2 & 2 \\
1 & 8 \end{array} \right).$$
Using the fact that $a(1,1) + b(0,2) = (a,a+2b)$, the transition matrix from the basis $\{(1,1),(0,2)\}$ to the standard basis $\{(1,0),(0,1)\}$ is
$$\left( \begin{array}{cc}
1 & 0 \\
1 & 2 \end{array} \right).$$
The matrix representation of $T$ with respect to the standard basis (for both the domain and co-domain) is then
$$\left( \begin{array}{cc}
2 & 2 \\
1 & 8 \end{array} \right)\left( \begin{array}{cc}
1 & 0 \\
1 & 2 \end{array} \right)^{-1} =
\frac{1}{2}\left( \begin{array}{cc}
2 & 2 \\
1 & 8 \end{array} \right)\left( \begin{array}{cc}
2 & 0 \\
-1 & 1 \end{array} \right) =
\left( \begin{array}{cc}
1 & 1 \\
-3 & 4 \end{array} \right).$$
Thus $T(a,b) = (a+b,-3a+4b).$
