<s>[INST] 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.
 [/INST] 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 $A$.
</s>