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.
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.