Similar matrices have the same characteristic equation.
Proof
Matrices
\[A\]
and \[B\]
are similar if there is an invertible matrix \[P\]
such that \[A=P^{-1}BP\]
.The characteristic equation of the matrix
\[A\]
is the determinant of the matrix \[A- \lambda I\]
\[\begin{equation} \begin{aligned} det(A- \lambda I) &= det(P^{-1}(B- \lambda I)P) \\ &= det(P^{-1}) det (B- \lambda I) det(P) \\ &= \frac{1}{det(P)} det (B- \lambda I) det(P) \\ &= det (B- \lambda I) \end{aligned} \end{equation}\]
If matrices have the same characteristic solution it does not follow that they are similar. Matrices
\[A= \left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right) \: B= \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right) \]
both have the characteristic equation \[(1- \lambda )^2\]
but these matrices are not similar.