if A and B are similar matrices then the eigenvalues are same. is the converse is true? why?
thank u
consider the two matrices:
$\displaystyle \begin{bmatrix}1&0\\0&1 \end{bmatrix},\ \begin{bmatrix}1&1\\0&1 \end{bmatrix}$
these both have the single eigenvalue 1, but have different Jordan normal forms (in fact both matrices ARE in Jordan normal form), so they cannot be similar.
$\displaystyle A=\begin{pmatrix}0&0\\ 0&0\end{pmatrix}$ and $\displaystyle B=\begin{pmatrix}0&1\\ 0&0\end{pmatrix}$ gives a counter-example, since both $\displaystyle A$ and $\displaystyle B$ have $\displaystyle 0$ as eigenvalue with multiplicity $\displaystyle 2$ but are not similar since the only matrix which is similar to $\displaystyle A$ is $\displaystyle A$.