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.
 A: 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 $A$.
A: Let $(e_1, e_2)$ a basisi corresponding to the first matrix:
$$
Ae_1= 2e_1 \\
Ae_2= - e_1 + 2e_2 $$
Consider another basis $(x_1, x_2)$. We want $Ax_2 = x_2$, let us take $x_2 = e_1$. Then look for $x_2 = ae_1 + be_2$.
Can you find a value of $(a,b)$ such as the matrix in the basis $(x_1, x_2)$ is the second one?
A: we can show that $A=\pmatrix{0&-1\\0&0}$ and $B= \pmatrix{0&0\\1&0}$ similar by explicitly displaying $$\pmatrix{1&0\\0&0}=AU = \pmatrix{0&-1\\0&0} \pmatrix{0&1\\-1&0} =  \pmatrix{0&1\\-1&0} \pmatrix{0&0\\1&0} = UB= \pmatrix{1&0\\0&0}.$$
now, $$AU=UB \implies (A+2I)U=U(B+2I) .$$ therefore $A+2I$ and $B+2I$ are similar.
