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

YCor
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Wei Wang
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  • http://en.wikipedia.org/wiki/Matrix_equivalence – Junkie May 28 '11 at 11:12
  • setting $t=0$ get A and B equivalent, but not similar! – Wei Wang May 28 '11 at 11:34
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    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 00:37
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    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
  • @Vladimir. You cannot avoid at least using polynomials, because the answer to the question whether two matrices are similar or not is given in terms of similarity invariant, which are polynomials. Whether you take or not polynomials of matrices or matrices of polynomials is a matter of taste. But it would look a bit conservative to avoid both. – Denis Serre Oct 19 '11 at 12:18
  • @DenisSerre Your proof still works if we replace matrix rings by arbitrary noncommutative rings. –  Nov 16 '19 at 21:03
  • The last paragraph is hopeless as written, since it seems to assume that $A,B$ have a common (characteristic) eigenvector, which for $A,B$ similar is usually not the case. – YCor Aug 25 '22 at 12:27

3 Answers3

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The beautiful module-theoretic proof given by @user20948 in his comment can be translated into a (long but reasonably nice) elementary proof using right evaluations of (noncommutative) polynomials. I have now expanded this proof at

Darij Grinberg, Similar matrices and equivalent polynomial matrices.

(Mostly written up to have a readable reference around next time I need the result in a class.)

darij grinberg
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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.

Amritanshu Prasad
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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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    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. –  Oct 16 '18 at 15:30
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    It should have been the cokernel of $T\otimes1_R-1_M\otimes T$, not the kernel. –  Oct 17 '18 at 09:45
  • @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. – Amritanshu Prasad Oct 17 '18 at 10:54
  • @FrankScience Thanks for your comment, I modified my answer. – Amritanshu Prasad Oct 17 '18 at 11:05
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    Did you think of replacing matrix rings by arbitrary noncommutative rings? Seemingly Denis' proof still works, I wonder whether this also works. So let us fix notations. Let $A$ be a ring and $a,b\in A$. If we have $(t-a)p(t)=q(t)(t-b)$, then we have a commutative diagram of right $A[t]$-modules ($t$ is a central element in $A[t]$), where morphisms are given by left multiplications by $t-a,p(t),q(t),t-b$ respectively. Passing to the cokernel of $t-a$ and $t-b$, we get a map $\phi$ of right $A[t]$-modules $A\to A$. It follows that $\phi(a)=b\phi(1)$. –  Nov 16 '19 at 21:26
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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.

Michael Renardy
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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
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    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