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.
 A: You’ve gone astray at the very end: $a(1,1)+b(0,2)=(a,a)+(0,2b)=(a,a+2b)$, so what you’ve calculated is $T(a,a+2b)$, not $T(a,b)$.
You know that $T(0,2)=(2,8)$, and $(0,1)=\frac12(0,2)$, so $T(0,1)=\frac12T(0,2)=\frac12(2,8)=(1,4)$. You also know that $(1,0)=(1,1)-(0,1)$, so $T(1,0)=T(1,1)-T(0,1)=(2,1)-(1,4)$. Finally, you know that $(a,b)=a(1,0)+b(0,1)$, so $T(a,b)=aT(1,0)+bT(0,1)$. Can you finish it from here?
A: We have 
$$(1,0)=(1,1)-\frac{1}{2}(0,2)\qquad\text{and} \qquad(0,1)=\frac{1}{2}(0,2).\tag{$1$}$$ 
Note that $(a,b)=a(1,0)+b(0,1)$. So 
$$T(a,b)=aT(1,0)+bT(0,1).$$
Now use the values of $T(1,1)$ and $T(0,2)$, and Equations $(1)$, to find $T(1,0)$ and $T(0,1)$, and simplify a bit.
A: 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 $T(a,b) = (a+b,-3a+4b).$
