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I need to prove the similarity between two 5x5 matrices. One of them is very simple (it has only one non-null element), but the other one isn't. For example:

\begin{equation} \mathbf{A}=\left(\begin{matrix}1 & 2 & 3 & 4 & 5 \cr 6 & 7 & 8 & 9 & 10 \cr 11 & 12 & 13 & 14 & 15 \cr 16 & 17 & 18 & 19 & 20 \cr 21 & 22 & 23 & 24 & 25 \end{matrix}\right) \end{equation} \begin{equation} \mathbf{B}=\left(\begin{matrix}1 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0\end{matrix}\right) \end{equation}

How does B make things easier? If I'd try to find the characteristic polynomial of A it wouldn't be so quick (or easy), so is there a better way?

PS. These two matrices are probably not similar, it was just to give an example. The question is really about the approach for solving something like this. Thanks.

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  • $\begingroup$ I think what you want is to compute the Jordan Cannonical Form. There are lots of questions about JCFs here on math.stackexchange, although not all of them go into the details of how to compute them. From a cursory look this answer seems useful: math.stackexchange.com/a/1570583/101420, but you can find many more using the search function on the site. $\endgroup$ – Vincent Oct 30 '17 at 13:40
  • $\begingroup$ First of all, thanks for the reply, but the thing is, I haven't studied the Jordan Cannonical Form (not yet anyway), but I'm supposed to know how to solve this. Can you think of any other way I could solve it? But again, thanks for the help. $\endgroup$ – lucast Oct 30 '17 at 13:50
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You know that $A$ and $B$ are similar implies that they have the same characteristic polynomial. If they have the same characteristic polynomial, they obviously have the same eigenvalues, trace, and determinant. The trace of $A$ is $65$, which is not equal to the trace of $B$, which is 1.

In general, one can easily compute the trace of a matrix, making it an efficient way to show that two matrices are not similar.

EDIT: I fixed a mistake, originally I said "if and only if", which is not true.

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  • $\begingroup$ It's true, computing the trace will be easy and I think it'll help me prove what I need to. Thank you for the help. $\endgroup$ – lucast Oct 31 '17 at 0:35
  • $\begingroup$ Glad to help, and welcome to stack exchange. If this answers your question, don't forget to accept it. Also, two matrices with the same trace may not be similar, so you can only use this trick to prove that two matrices are NOT similar. $\endgroup$ – amarney Oct 31 '17 at 12:35

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