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

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setting $t=0$ get A and B equivalent, but not similar! – Wei Wang May 28 '11 at 11:34
See my book Matrices, published as a Springer-Verlag GTM 216. 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. – Denis Serre May 28 '11 at 13:07
Denis: Thank you! It is clearer. However, is there any geometric proof (by showing some properties of eigenvalues, certain subspaces, etc.)? – Wei Wang May 29 '11 at 0:37
In my opinion, if you want to study similarity, Jordan forms etc. in a geometric way, there are great ways to do it without polynomial matrices. One of the main points of using polynomial matrices is to demonstrate that these questions of linear algebra are just a shadow of general (rather abstract) results on modules over principal ideal domains. – Vladimir Dotsenko Jun 11 '11 at 14:56

Suppose there are matrices P and Q such that P(tI-A)Q=tI-B for all t. Then we conclude that PQ=I, PAQ=B. Or am I missing something?

If P and Q are allowed to depend on t, then all we can conclude is that tI-A and tI-B have the same rank for every t. This is not enough to make A and B similar.

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Michael: $P$ and $Q$ have polynomial entries. Therefore it is not straightforward that $PQ=I$ and $PAQ=B$. – Denis Serre May 28 '11 at 13:04
Thank you for the clarification. Now I understand the problem. It seems that fact P and Q are polynomials must indeed be used, since the result fails if we allow a general dependence on t. On the other hand, the partial results outlined in the original posting do not depend on P and Q being polynomials. – Michael Renardy May 28 '11 at 14:24

I do not have Denis Serre's book, so I do not know if this answer is the same as the one in his book. It is certainly very simple. It is not particularly geometric, and may alas be pretty close to proofs in the literature.

As usual, take $M^A$ and $M^B$ to be the $k[t]$-modules with underlying space $k^n$, where $t$ acts by $A$ and $B$ respectively. Then $A$ is similar to $B$ if and only if $M^A$ and $M^B$ are isomorphic as $k[t]$-modules. As $k[t]$ modules $M^A$ and $M^B$ are both generated by the coordinate vectors $e_1,\dotsc,e_n$, and given by relations (in matrix form)

$\begin{pmatrix} e_1 & e_2 & \cdots & e_n \end{pmatrix}(A-It)=0$


$\begin{pmatrix} e_1 & e_2 & \cdots & e_n \end{pmatrix}(B-It)=0$

In general, modules over Euclidean domains given by matrices of relations are isomorphic if and only if the matrices have the same Smith normal form. Applying this to the present instance gives the condition that $A-tI$ and $B-tI$ are equivalent.

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thanks. but it's for freshmen, and they certainly don't know what are modules. – Wei Wang May 11 '12 at 12:45
Did you find an elementary AND simple answer, please post ir here; I too would like to know. – Amritanshu Prasad May 11 '12 at 16:39

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