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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.

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

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  • $\begingroup$ Nice shortcut!${}{}{}$ $\endgroup$ – rschwieb Nov 29 '17 at 3:27
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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.

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  • $\begingroup$ I suspect that if OP were conversant with Jordan normal form, then OP would not have posted the question. $\endgroup$ – Gerry Myerson Nov 29 '17 at 2:46
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    $\begingroup$ @GerryMyerson Then that is exactly what op need to learn. Good thing I explained it in my answer. $\endgroup$ – Rene Schipperus Nov 29 '17 at 2:48
  • $\begingroup$ So, you consider that your answer explains how to distinguish between matrices that are, and are not, in Jordan normal form? $\endgroup$ – Gerry Myerson Nov 29 '17 at 2:57
  • $\begingroup$ +1 Hooray a totally different approach via a different insight! God bless variety of answers! It’s great that our users here don’t discourage such answers because the original poster merely might not be totally ready for them yet. And it’s also wonderful that they never made prequels to the original Star Wars movies... $\endgroup$ – rschwieb Nov 29 '17 at 3:35
  • $\begingroup$ @rschwieb Thanks for the sarcasm, this site need more of that. $\endgroup$ – Rene Schipperus Nov 29 '17 at 3:41

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