How to find $3\times3$ matrices that satisfy the matrix equation $A^2=I_3$? How to find $3\times3$ matrices that satisfy the matrix equation $A^2=I_3$? Can anyone please show me steps to do this question?
 A: All of the matrices you want can be obtained by taking $A$ of the following form:
$\begin{pmatrix}
a_1 & 0 & 0 \\
0 & a_2 & 0 \\
0 & 0 & a_3\\
\end{pmatrix}$
where $a_1,a_2$ and $a_3$ are $1$ or $-1$.
And then taking $B$ an invertible matrix (possibly with complex entries) and considering the matrix $BAB^{-1}$
A: Geometrically, the four (classes of) solutions will be:


*

*the identity ($I$);

*reflection through the origin ($-I$);

*reflection through any plane through the origin;

*rotation of $180^\circ$ about any axis through the origin.
Proof outline: The equation
$$
A^2 = I_3
$$
factors as
$$
(A - I_3)(A+I_3) = 0
$$
Now if we pick a basis so that $A$ is in Jordan normal form, we have
$$
A = \begin{bmatrix} x & a & 0 \\ 0 & y & b \\ 0 & 0 & z\end{bmatrix}
$$
so
$$
\begin{bmatrix} x-1 & a & 0 \\ 0 & y-1 & b \\ 0 & 0 & z-1\end{bmatrix}
\begin{bmatrix} x+1 & a & 0 \\ 0 & y+1 & b \\ 0 & 0 & z+1\end{bmatrix}
=
\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}.
$$
By comparing entries, you should be able to show
that:


*

*$x = \pm 1$, $y = \pm 1$, $z = \pm 1$

*$ab = 0$

*$a(x+y) = 0$, $b(y+z) = 0$
Use these equations (and don't forget it's in Jordan normal form!) to narrow down to it being diagonal. There should then be just $4$ cases, up to permuting the Jordan blocks.
For each of these possibilities for $A$, taking an arbitrary invertible $S$,
$$
S A S^{-1}
$$
will give you all possible solutions in any basis.
A: Note that $x^2+1=0$ is a annihilating polynomial of $A$. Now minimal polynomial of a matrix divides a annihilating polynomial.
So the possible choices are $x=-1,x=1,x^2+1$.
Hence the matrices in first two cases will be $A=I,A=-I$.
Consider the third case.Since the minimal polynomial and characteristic polynomial have the same roots so characteristic polynomial will be $(x-1)^2(x+1),(x+1)^2(x-1)$.
Hence the possible choices for $A$ will be :
\begin{bmatrix} 1&0 &0\\0&1&0\\0&0&-1\end{bmatrix}
or \begin{bmatrix} -1&0 &0\\0&-1&0\\0&0&1\end{bmatrix}
