Similarity between $A$ and $B (2\times 2)$. I am trying to find out whether $A = \begin{bmatrix}
    3 & 1 \\
    0 & 2 \\
\end{bmatrix}$ and $B = \begin{bmatrix}
    4 & 1 \\
    -2 & 1 \\
\end{bmatrix}$ are similar. Both matrices have the same trace, same rank, same determinant and the same eigenvalues. So I think they are similar. The eigenvalues are $3, 2$. Now I'm not sure how to find if they are indeed similar. 
I am trying to find if there is a basis for $\mathbb{R^2}$ that contains only eigenvectors but when trying to solve $(3 I - A)v = 0$ there is no solution... What should I do from here?
 A: Similarity is an equivalence relation (prove it). So if $A$ and $B$ are similar to the same matrix, then they are similar.
Since the eigenvalues of $A$ and $B$ are the same (and distinct), they both are similar to
$$
\begin{bmatrix}
3 & 0 \\
0 & 2
\end{bmatrix}
$$
so they're similar. No other computation is needed.
A: Of course there is a solution for $(3I-A)v=0$: all vectors $(x,0) $. 
In general, if an $n\times n $ matrix has $n $ distinct eigenvalues,  then it is diagonalizable, so your matrices are indeed similar.
A: Sketch:
(1) Find the eigenvectors for each of the matrices.
(2) Consider the map $M$ that takes the eigenvectors for $A$ to the eigenvectors for $B$ with corresponding eigenvalues.
(3) Consider $M^{-1}BM$.
Why does this work: Let $\lambda_1\not=\lambda_2$ be the eigenvalues of $A$ and $B$.  Let $v_1,v_2$ be the (corresponding) eigenvectors of $A$ and $w_1,w_2$ be the (corresponding) eigenvectors of $B$.  Then $Av_i=\lambda_iv_i$.  On the other hand, $M^{-1}BMv_i=M^{-1}Bw_i=M^{-1}\lambda_iw_i=\lambda_iM^{-1}w_i=\lambda_iv_i$.  Basically $A$ and $B$ act the same way on the corresponding eigenvectors, so by mapping one to the other, they act the same way.
A: If $A=PDP^{-1}$, and $B=QDQ^{-1}$, where $D$ is the diagonal matrix with diagonal entries $2$ and $3$, then $A=PQ^{-1}BQP^{-1}$, and you have done.
