$$A=\begin{bmatrix} 1 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} \hspace{1cm} B=\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{bmatrix} \hspace{1cm} C=\begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

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?

  • 1
    $\begingroup$ Some holes in your logic: 1. Similar matrices have the same eigenvalues, but the converse need not be true. 2. Your justification of the claim "$A$ and $C$ are diagonalizable" is incorrect; $C$ is actually not diagonalizable. $\endgroup$ – angryavian May 23 '18 at 16:19
  • $\begingroup$ @Daphne Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… $\endgroup$ – user Jun 22 '18 at 20:14


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 $=1$).


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