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how can we find the number of matrices with real entries of size $9 \times 9$ (up to similarity) such that $A^{2}=I$?

I first thought about the following:

Notice $A$ satisfies the polynomial $f(t)=t^{2}-1$ hence its minimal polynomial divides $(t-1)(t+1)$.

So its characteristic polynomial is of the form $p(t)=(t-1)^r(t+1)^j$ where $r+j = 9$, right? Then I'm not sure what to do, I tried to consider the rational canonical form but in order to do this we need to know the minimal polynomial right? because in the rational canonical form the last term in the array is exactly the minimal polynomial, how to find it?

Can you please help?

Martin Sleziak
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user10
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3 Answers3

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The eigenvalues are indeed $\pm 1$, so the equation $A^2=I$ is solved exactly by all $A$ of the form $$ A = C D C^{-1} $$ where $C$ is an arbitrary non-singular $9\times 9$ matrix and $D$ is an arbitrary diagonal matrix with $r$ entries $+1$ and $j$ entries $-1$, $r+j=9$. This set of the solution has several disconnected components labeled by the labels $(r,j)$. Each component has the dimension $80-36=44$, I guess, because among the $80$ a priori possible generators of $SL(9)$, the generators of $SO(r,j)$ don't change the matrix.

Luboš Motl
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  • Motl: thanks, can you please explain more your reasoning? – user10 Jun 09 '11 at 15:24
  • Hi, every $9\times 9$ matrix $A$ may be brought into the standard form $A=CDC^{-1}$ for a $D$ which is either diagonal or has the Jordan blocks on the diagonal. That's a basic result in algebra. In this form, $A^2 = CDC^{-1}CDC^{-1} = CD^2 C^{-1}$. It should be equal to $I = CC^{-1}$ which implies $D^2=I$. So $D$ has to have $\pm 1$ eigenvalue and one may check that the nondiagonal Jordan blocks would fail to produce $D^2=1$, too. – Luboš Motl Jun 09 '11 at 15:30
  • @ Lubos : The question was to count the matrices up to similarity, so your first sentence already give the full answer : 10 (not that what you write afterwards is wrong or anything). Also note that since $X^2-1$ has simple roots, it is fairly easy to prove that $A$ is diagonalizable without using the full strengh of Jordan decomposition (basically just prove the $\ker(A-I)$ and $\ker(A+I)$ span the whole space). – Joel Cohen Jun 09 '11 at 16:40
  • @JoelCohen: How could we extend the count for n>2 ? Characteristic polynomial will not necessarily split, at least not over the Reals. – gary Jun 23 '18 at 22:30
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As you note, the minimal polynomial divides $(t-1)(t+1)$. Since the minimal polynomial splits and is square free, that means that the matrix is necessarily diagonalizable. Therefore, you want a diagonalizable matrix with eigenvalues $-1$ and/or $1$. Just pick how many times $1$ is an eigenvalue (from $0$ through $9$) to get all similarity types.

Arturo Magidin
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    Here's an elementary proof that $A$ is diagonalizable with eigenvalues $1$ and $-1$ : basically, it's a symmetry. First of all, check that $\ker(A-I) \cap \ker(A+I) = (0)$ (holds for any matrix). And if $x \in \mathbb{R}^9$ then one can write

    $$x = \frac{x+Ax}{2} + \frac{x-Ax}{2}$$

    and we have $\frac{x+Ax}{2} \in \ker(A-I)$, and $\frac{x-Ax}{2} \in \ker(A+I)$. So

    $$\mathbb{R}^9 = \ker(A-I) \oplus \ker(A+I)$$

     – Joel Cohen Jun 09 '11 at 17:02
  • @Arturo Magidin: Thank you, can you please explain more? (the line "how many times 1 is an eigenvalue (from 0 through 9) to get all similarity types" confuses me. – user10 Jun 09 '11 at 20:07
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    @user10: Since the matrix is diagonalizable, the matrix will be similar to a diagonal matrix with $1$s and $0$s in the diagonal. The order of the diagonal entries doesn't matter, because a diagonal matrix obtained by shuffling the diagonal entries of another diagonal matrix is similar to the original one. So you can assume that all the $0$s come first, and all the $1$s come later. At this point, the only thing you need to decide is how many $1$s there will be; different quantities of $1$s yield nonsimilar matrices. How many $1$s can you have? Any number from $0$ to $9$. – Arturo Magidin Jun 09 '11 at 20:09
  • @Arturo Magidin: Thanks. How would this change if instead we work over a finite field, say $\mathbb{F}{3}$ or $\mathbb{F}{5}$? – user10 Jun 09 '11 at 20:47
  • @user10: It wouldn't. The minimal polynomial splits and has no repeated roots, so the matrix is diagonalizable, regardless of what field you are working over. – Arturo Magidin Jun 09 '11 at 21:00
  • @Arturo : Actually, to be precise, this does not works in characteristic $2$ (for example in $\mathbb{F}_2$). In this case there, I think the number of similarity classes is $30$, the number of partition of $9$ (because of the Jordan form). – Joel Cohen Jun 09 '11 at 22:26
  • @Joel Cohen: is it possible to solve this without Jordan forms? – user10 Jun 09 '11 at 22:40
  • @user10 : I think I was mistaken with the number 30: I was counting (similarity classes of) nilpotent matrices, but not necessarily nilpotent of order ≤2 (which would give you the number of partitions of 9 into "ones" and "twos", that is 5 I believe). – Joel Cohen Jun 09 '11 at 23:37
  • @user10: Here's an ad hoc proof (Jordan in disguise). You need to count matrices with $N^2 = 0$. I claim that such matrices are similar if and only if they have same rank (which gives $5$ classes because $\textrm{Im}(N) \subset \ker(N)$ gives $2\ \textrm{rk}(N) \le 9$): picking a supplementary of $\ker(N)$, $N$ is similar to

    $$\left(\begin{matrix}0&X\0&0\end{matrix}\right)$$

    Where $X$ is rectangular of rank $r$. There exist $P$, $G$ such that $PXQ^{-1}=J_r=(\delta_{ij})$. Conjugating by $\textrm{diag}(P,Q)$, $N$ is similar to

    $$\left(\begin{matrix}0&J_r\0&0\end{matrix}\right)$$

     – Joel Cohen Jun 10 '11 at 01:03
  • @Joel: Quite so, because in characteristic 2 the minimal polynomial is not square free, since $-1=1$. But your computations are incorrect, because the minimal polynomial is still a divisor $(t-1)^2$, so the largest possible Jordan block is $2\times 2$. In that case, the minimal polynomial divides $(t+1)^2$; if the minimal polynomial is $t-1$, then you get exactly one similarity type. If the minimal polynomial is $(t+1)^2$, then you get one similarity type for each partition of 9 that has at least one part of size 2 and no larger part: 2+2+2+2+1, 2+2+2+1+1+1, 2+2+1+1+1+1+1, and 2+1+1+1+1+1+1+1. – Arturo Magidin Jun 10 '11 at 03:05
  • @Joel: So the total is five similarity types. Which I now see you corrected later. – Arturo Magidin Jun 10 '11 at 03:06
  • @ArturoMagidin: But if you have 0s in the diagonalized matrix, you will not satisfy $A^2=I$. Am I missing something? Moreover, can you suggest how to go about finding solutions to $A^n =Id $, please? – gary Jun 23 '18 at 22:09
  • @gary: Who said anhything about $0$s in the diagonal? The diagonalized matrix will have zeroes everywhere but the diagonal, and the diagonal will contain only $1$s and $-1$s. Where are you getting $0$s from? And, no, I will not go through a full solution seven years later, for your benefit. You have all the information you need. – Arturo Magidin Jun 24 '18 at 01:10
  • @Joel I am trying to generalize to solutions of $A^n =Id $ and I just want a hint . I know this relates to the minimal poly of A^n -Id splitting into linear factors. I know the eigenvalues (when/if a solution exists) are the n-th roots of unity. Maybe we can induct on the trick $SD^2S^{-1} =Id =SS^{-1}$ implies ( by Lubos' post below) $D^2 =\pm Id $ then we have $SD^nS^{-1} =Id =SS^{-1} \rightarrow D^n =Id $ But this assumes A is diagonalizable. I have thought that, e.g., rotations R by amounts not conmensurable to $2 \pi$ will not have a solution to $R^n =Id$. Any ideas, please – gary Jun 27 '18 at 18:34
  • @gary: Then ask the question in the site, not buried in the comments of an answer from 7 years ago from someone whose profile states that he does not want to resume his active participation here. – Arturo Magidin Jun 27 '18 at 19:56
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As pointed out in another answer, $A$ has to have Jordan blocks $A_i$ of the form $\pm I +N$ where $N$ is a matrix with a certain number of $1$'s in the first off-diagonal and the rest zeros. The matrix $N^2$ has a certain number of $1$'s in the second off-diagonal and the rest zeros. It follows that the condition $(A_i)^2 = I$, i.e., $I\pm 2 N+ N^2 =I$ implies $N=0$. Therefore $A$ is similar to a diagonal matrix with all diagonal entries $1$ or $-1$. There are 10 different types of such matrices.

Christian Blatter
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