<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] 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.
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