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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?

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3 Answers 3

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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.

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  • $\begingroup$ Motl: thanks, can you please explain more your reasoning? $\endgroup$
    – user10
    Jun 9, 2011 at 15:24
  • $\begingroup$ 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. $\endgroup$ Jun 9, 2011 at 15:30
  • $\begingroup$ @ 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). $\endgroup$
    – Joel Cohen
    Jun 9, 2011 at 16:40
  • $\begingroup$ @JoelCohen: How could we extend the count for n>2 ? Characteristic polynomial will not necessarily split, at least not over the Reals. $\endgroup$
    – gary
    Jun 23, 2018 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.

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    $\begingroup$ 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 $$\mathbb{R}^9 = \ker(A-I) \oplus \ker(A+I)$$ $\endgroup$
    – Joel Cohen
    Jun 9, 2011 at 17:02
  • $\begingroup$ @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. $\endgroup$
    – user10
    Jun 9, 2011 at 20:07
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    $\begingroup$ @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 $9$. $\endgroup$ Jun 9, 2011 at 20:09
  • $\begingroup$ @Arturo Magidin: Thanks. How would this change if instead we work over a finite field, say $\mathbb{F}_{3}$ or $\mathbb{F}_{5}$? $\endgroup$
    – user10
    Jun 9, 2011 at 20:47
  • $\begingroup$ @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. $\endgroup$ Jun 9, 2011 at 21:00
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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.

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