Let $T:\mathbb{R}^2\rightarrow \mathbb{R}^2$ such that:
$$T\left( {\matrix{ x \cr y \cr } } \right) = \left( {\matrix{ 2 & 1 \cr 3 & 4 \cr } } \right)\left( {\matrix{ x \cr y \cr } } \right)$$
Show that there's no ordered-basis, $E$ such that:
$${\left[ T \right]_E} = \left( {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right)$$
I assumed that there's such basis (contains two vectors) and applied $T$ on them. Then, compared $T(v_1) = (1 \space 0)^T$ and $T(v_2) = (0 \space 1)^T$
and I did got a basis answering the conditions above. No contradiction.
What did I miss?
When you write $T(v_1) = (1\ 0)^T$, you're writing the coordinates of the vector $T(v_1)$ in the standard basis. But the columns of $[T]_E$ record the coordinates of $T(v_1)$ and $T(v_2)$ in the $\{v_1,v_2\}$ basis. That's why your solution to the first equation doesn't work in the problem as stated.