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

I'm teaching linear algebra and I'm encountering 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 $(kI - B)^n x = 0$ ...