I wanted to know Is there exist 2 nonsimilar matrices with all algebraic properties same?
I think there exists such pair as otherwise there we necessary sufficient condition of diagonalisibily using above.
Please Help me to construct an example
I can construct 2 nonsimilar matrices with same minimal polynomial,characteristics polynomial eigenvalue. Example $$A=\begin{pmatrix} 1 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 0 & 0 & 0 & 1\end{pmatrix}$$,
$$B= \begin{pmatrix} 1 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 1\end{pmatrix},$$
Here Eigen values of both are A and B are 1 Minimal Polynomial is $(x-1)^3$ and Char Polynomial is $(x-1)^7$ Eigenvectros of A is $\{e_1,e_4,e_6\}$ and that of B $\{e_1,e_4,e_7\}$
But I do not know how to convert above matrices by permutation to have same eigenvectors Any Help will be appreciated
I think Eric is hinting at something like this: $$\pmatrix{0&1&&&&&\cr&0&1&&&&\cr&&0&0&&&\cr&&&0&1&&\cr&&&&0&0&\cr&&&&&0&1\cr&&&&&&0\cr}$$ and $$\pmatrix{0&1&&&&&\cr&0&1&&&&\cr&&0&0&&&\cr&&&0&1&&\cr&&&&0&0&1\cr&&&&&0&0\cr&&&&&&0\cr}$$
If I've done it right, both have characteristic polynomial $t^7$, minimal polynomial $t^3$, and eigenvectors $(1,0,0,0,0,0,0)$, $(0,0,0,1,0,0,0)$, and $(0,0,0,0,0,1,0)$, but they aren't similar.