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 $(kI - B)^n x = 0$ ...

Matrices, published as a Springer-Verlag GTM216. In the 2nd edition, this is Theorem 9.5/9.6, in Section 9.3. The proof takes one page. It is a beautiful piece of mathematics, to my taste. $\endgroup$ – Denis Serre May 28 '11 at 13:07