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

  • $\begingroup$ en.wikipedia.org/wiki/Matrix_equivalence $\endgroup$
    – Junkie
    May 28, 2011 at 11:12
  • $\begingroup$ setting $t=0$ get A and B equivalent, but not similar! $\endgroup$
    – Wei Wang
    May 28, 2011 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. $\endgroup$ May 28, 2011 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, 2011 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. $\endgroup$ Jun 11, 2011 at 14:56

3 Answers 3


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.)


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)

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


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

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.

  • $\begingroup$ thanks. but it's for freshmen, and they certainly don't know what are modules. $\endgroup$
    – Wei Wang
    May 11, 2012 at 12:45
  • $\begingroup$ Did you find an elementary AND simple answer, please post ir here; I too would like to know. $\endgroup$ May 11, 2012 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. $\endgroup$
    – user20948
    Oct 16, 2018 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. $\endgroup$
    – user20948
    Oct 17, 2018 at 9:45
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    $\begingroup$ 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 $\phi(a)=b\phi(1)$. $\endgroup$
    – user20948
    Nov 16, 2019 at 21:26

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.

  • 1
    $\begingroup$ Michael: $P$ and $Q$ have polynomial entries. Therefore it is not straightforward that $PQ=I$ and $PAQ=B$. $\endgroup$ May 28, 2011 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. $\endgroup$ May 28, 2011 at 14:24

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