# On similar matrices.

I am searching two matrices such that both of the matrices have same characteristic polynomial, same minimal polynomial, same rank, same trace, same determinant, same algebraic and geometric multiplicity of corresponding eigenvalues but still are not similar. Now we know that two matrix are similar if they have the same Jordan canonical form except of some arrangements of Jordan blocks. On basis of this result i am giving two matrices having same characteristic polynomial $(x^{7})$, same minimal polynomial $(x^{3})$, same rank $(4)$, same trace $(0),$ same determinant $(0)$ and same algebraic and geometric multiplicity i.e. $7$ and $3$ respectively of the eigenvalue $0.$ The first matrix has only one Jordan block of order $3$, but the second one has $2$ Jordan blocks of order $3.$ Can i say that the two matrices are are not similar?

And these are our required matrices as discussed above? Please explain. Thanks in advance.

\begin{bmatrix} 0 &1 &0&0&0&0& 0 \\ 0 &0 &1&0&0&0& 0 \\ 0 &0 &0&0&0&0& 0 \\ 0 &0 &0&0&1&0& 0\\ 0 &0 &0&0&0&0& 0\\ 0 &0 &0&0&0&0& 1\\ 0 &0 &0&0&0&0& 0 \end{bmatrix}

\begin{bmatrix} 0 &1 &0&0&0&0& 0 \\ 0 &0 &1&0&0&0& 0 \\ 0 &0 &0&0&0&0& 0 \\ 0 &0 &0&0&1&0& 0\\ 0 &0 &0&0&0&1& 0\\ 0 &0 &0&0&0&0& 0\\ 0 &0 &0&0&0&0& 0 \end{bmatrix}

• Sorry, ignore what I just wrote. I was wrong. – 5xum Sep 13 '15 at 17:29
• The matrices (although having same this and that) are not similar: one Jordan form cannot be obtained as a permutation of the Jordan blocks of the other. – Algebraic Pavel Sep 13 '15 at 21:17
• its mean my discussion is right? – neelkanth Sep 14 '15 at 1:46