Prove that matrices $\tiny\begin{pmatrix} 2&-1 \\ 0&2 \\ \end{pmatrix},\begin{pmatrix} 2& 0 \\ 1&2 \\ \end{pmatrix} $ are similar. Error in my method? Show that the matrix 
$$\begin{pmatrix} 2&-1 \\ 0&2 \\ \end{pmatrix}  $$ is similar to a triangular matrix of the form  $$ \begin{pmatrix} \lambda& 0 \\ 1&\lambda \\ \end{pmatrix}  $$ where $\lambda$ is an eigen value.
Attempt:
The eigenvalues of the matrix $\begin{pmatrix} 2&-1 \\ 0&2 \\
\end{pmatrix}  $ are clearly $2,2$.
Hence, we need to show that the matrix $\begin{pmatrix} 2&-1 \\0&2 \\
\end{pmatrix}  $ is similar to $\begin{pmatrix}2&0 \\1&2 \\\end{pmatrix}  $

Now, we know that two matrices are similar if and only if they represent the same linear transformation.

But: for any two dimensional vector $(x ~~y)^T : $
$\begin{pmatrix} 2&-1 \\ 0&2 \\ \end{pmatrix} \begin{pmatrix}x \\y \\
\end{pmatrix} =  \begin{pmatrix} 2x-y \\ 2y \\ \end{pmatrix}  $ 
And 
$\begin{pmatrix} 2&0 \\1&2 \\\end{pmatrix} \begin{pmatrix}x \\ y \\\end{pmatrix} =  \begin{pmatrix} 2x \\x+2y \\\end{pmatrix}  $ 
Clearly, $\begin{pmatrix} 2x \\x+2y \\\end{pmatrix}  $  and $\begin{pmatrix}
2x-y \\2y \\ \end{pmatrix}  $ don't represent the same linear transformation.
So, how can these matrices be similar? 
What could be the fault in my reasoning?
Thank you very much for your help in this regard.

Call $A$ the first matrix, written with respect to a basis $e_{1}, e_{2}$ and $B$ the second, written with respect to a basis $f_{1}, f_{2}$. 
Note that $A$ fixes $e_{1}$, and $B$ fixes $f_{2}$. Moreover, $A e_{2} = 2 e_{2} - e_{1}$. Can you find a number $a$ so that for the vector $f_{1} + a f_{2}$ one has 
$$
B (f_{1} + a f_{2}) = 2 (f_{1} + a f_{2}) - f_{2}?
$$
If you can do that, the matrix of $B$ with respect to the basis $f_{2}, f_{1} + a f_{2}$ will be $A$.
