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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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  • $\begingroup$ en.wikipedia.org/wiki/Matrix_equivalence $\endgroup$ – Junkie May 28 '11 at 11:12
  • $\begingroup$ setting $t=0$ get A and B equivalent, but not similar! $\endgroup$ – Wei Wang May 28 '11 at 11:34
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    $\begingroup$ 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. $\endgroup$ – Denis Serre May 28 '11 at 13:07
  • $\begingroup$ Denis: Thank you! It is clearer. However, is there any geometric proof (by showing some properties of eigenvalues, certain subspaces, etc.)? $\endgroup$ – Wei Wang May 29 '11 at 0:37
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    $\begingroup$ 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. $\endgroup$ – Vladimir Dotsenko Jun 11 '11 at 14:56
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This proof is different from the one in Denis Serre's book.

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)

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

and

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

In general, modules given by matrices of relations are isomorphic if and only if the relation matrices are equivalent.

Thus $A$ and $B$ are similar if and only if $A-tI$ and $B-tI$ are equivalent.

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  • $\begingroup$ thanks. but it's for freshmen, and they certainly don't know what are modules. $\endgroup$ – Wei Wang May 11 '12 at 12:45
  • $\begingroup$ Did you find an elementary AND simple answer, please post ir here; I too would like to know. $\endgroup$ – Amritanshu Prasad May 11 '12 at 16:39
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    $\begingroup$ Sorry for a late comment, but I don't understand where you use Smith normal form. Let $R=k[T]$, and I will write $M,N$ instead of $M_A,M_B$ in your post. What you have shown is that, $A,B$ are similar iff $M\cong N$ as $R$-modules. The presentation for $M$ you gave is essentially saying that $M$ is $R$-isomorphic to the kernel of the $R$-endomorphism $T\otimes1_R-1_M\otimes T$ on $M\otimes_kR$, and a similar description for $N$ holds. If $A-T,B-T$ are equivalent, then these kernels are isomorphic (as $R$-modules). These arguments are quite formal, not dependent on the fact that $R$ is a PID. $\endgroup$ – Yai0Phah Oct 16 '18 at 15:30
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    $\begingroup$ It should have been the cokernel of $T\otimes1_R-1_M\otimes T$, not the kernel. $\endgroup$ – Yai0Phah Oct 17 '18 at 9:45
  • $\begingroup$ @FrankScience Your'e right. Matrices of relations give isomorphic modules if and only if they are equivalent. There is no need to use Smith form. However, from the Smith normal form of $A-It$ you will be able to read off the rational canonical form of A, and if you first separate the primary parts, and then do Smith form, you can read off the Jordan canonical form. $\endgroup$ – Amritanshu Prasad Oct 17 '18 at 10:54
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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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    $\begingroup$ Michael: $P$ and $Q$ have polynomial entries. Therefore it is not straightforward that $PQ=I$ and $PAQ=B$. $\endgroup$ – Denis Serre May 28 '11 at 13:04
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    $\begingroup$ 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. $\endgroup$ – Michael Renardy May 28 '11 at 14:24

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