# Are these two $3\times 3$ matrices similar?

Is $A = \begin{bmatrix} 1&1&0\\ 0&1&0\\ 0&0&1\\ \end{bmatrix}$ and $B = \begin{bmatrix} 1&1&0\\ 0&1&1\\ 0&0&1\\ \end{bmatrix}$ similar? Please justify your answer.

So far what I've done is to check rank, det, trace, and characteristic polynomial to maybe disprove it but all of them are the same so I'm kinda stuck.

• How about minimal polynomial? – Gerry Myerson Nov 29 '17 at 2:18
• or its rational form? – janmarqz Nov 29 '17 at 2:19
• Any thoughts about the answers that have been posted, confused? – Gerry Myerson Nov 30 '17 at 22:27

If $A$ and $B$ are similar, $A-I$ and $B-I$ must be similar too, which means that they must have the same rank. This is not the case here, thus $A$ and $B$ are not similar.
• Nice shortcut!${}{}{}$ – rschwieb Nov 29 '17 at 3:27
These are matrices in Jordan normal form. This is a representative of the similarity class. Thus these matrices are not similar. For example $A$ has two independent eigenvectors to the eigenvalue $1$ whereas $B$ has only one.