<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] 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?
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