Which of these matrices are similar? $$A=\begin{bmatrix}
    1 &amp; 1 &amp; 0 \\
    0 &amp; 2 &amp; 0 \\
    0 &amp; 0 &amp; 2
\end{bmatrix}  
\hspace{1cm} B=\begin{bmatrix}
    2 &amp; 0 &amp; 0 \\
    0 &amp; 2 &amp; 0 \\
    0 &amp; 0 &amp; 1
\end{bmatrix}
\hspace{1cm} C=\begin{bmatrix}
    2 &amp; 1 &amp; 0 \\
    0 &amp; 2 &amp; 0 \\
    0 &amp; 0 &amp; 1
\end{bmatrix}$$

I would argue as follows:

Since all matrices are triagular, each has the same set of eigenvalues. As $B$ is diagonal, we can say that both $A$ and $C$ are diagonalizable, with their diagonal matrices being $B$. It follows that $A$ and $C$ are similar to $B$. By transitivity, they are also similar to each other. Hence all matrices are similar.

However, I was told that the order of the eigenvalues (it differs among the matrices) might invalidate this reasoning, but I don't see how that could be true, given that we may asign the order of the eigenvalues arbitrarily when diagonalizing a matrix.

On second though, I could just use the theorem that says that similar matrices have the same eigenvalues, but would that line of reasoning above still be valid?
HINT

Matrices $A$ and $B$ are similar indeed $B$ is the diagonal form of $A$.

Matrix $C$ is a Jordan canonical form (i.e. eigenvalues $2$ has geometric multiplicity $=1$).
HINT

Matrices $A$ and $B$ are similar indeed $B$ is the diagonal form of $A$.

Matrix $C$ is a Jordan canonical form (i.e. eigenvalues $2$ has geometric multiplicity $=1$).