I m teaching linear algebra and encounter this theorem:
two matrices A and B are similar iff tI - A and tI - B are equivalent (as polynomial matrices), where I is the unit matrix.
The proof that I learned and found on all available textbooks is very tricky (to me). So I try to get a more intuitive proof, but end up with the following:
if tI - A and tI - B are equivalent, then A and B have same eigenvalues, and the corresponding eigenvector subspaces are of same dimensions etc.
So, can we move forward in this direction? e.g., if k is an eigenvalue for both A and B and $(kI - A)^n x = 0$ then $(kI - B)^n x = 0$ ...