Is $A^2=I \implies A=\pm I$ necessarily true? $A$ is $n\times n$ matrix.
How to prove whether  it is true or false $$A^2=I \implies A=\pm I$$
I was trying on $2\times 2$ case...multiplying general entries and then equating them to the identity requirements...but it is not a proof...
 A: Any time that you get back to where you started by doing an operation twice, you have the situation $A^2=I$. Such operations are called involutions. So, in the plane, reflection about any axis that passes through the origin will have this property. Turning your shirt inside-out is probably not a linear operation, so it won’t have a matrix, but it is an involution. These operations are everywhere, even though you can’t unring a bell.
A: Hint:
$$A=\begin{pmatrix}0&-1\\-1&\;0\end{pmatrix}$$
A: Hint: It is false. Complete the following matrix suitably: $\begin{bmatrix}\clubsuit & 0\\0 &\spadesuit\end{bmatrix}$.

More generally, given natural numbers $m,n$ and $A\in \Bbb C^{n\times }$ such that $A^m=I_n$, it is true that the eigenvalues of $A$ are $m$ roots of $1$, (not necessarily all of them).
Furthermore $A$ is necessarily diagonalizable.
This facts allows one to easily build counter examples for the statement $A^m=I_n\implies A=\pm I_n$, if $m\ge 2$. For instance, if $m=n$, let $A=\operatorname{diag}(e^{\large{2\pi i/m}}, e^{\large{4\pi i/m}}, \ldots,e^{\large{2m\pi i/m}})$.
A: For a slightly more geometric approach, consider the definition of a matrix as a linear transformation; a mapping $A : \mathbb{R}^n \rightarrow \mathbb{R}^n$ which fixes the origin (and has other properties: in general, $A(au+bv) = aA(u) + bA(v)$, where $u, v$ are vectors and $a, b$ are scalars).  Since we are working with a $2 \times 2$ matrix, $n = 2$, and we are mapping two dimensional space (the plane) onto itself.
A reflection across the line $y = ax$ is a linear transformation in two dimensional space, and can be written as
$$A = \left[ \begin{array} {cc}
\frac{a^2 - 1}{a^2 + 1} & \frac{2a}{a^2 + 1} \\
\frac{2a}{a^2+1} & \frac{1-a^2}{a^2+1}
\end{array}\right].$$
Clearly, applying a reflection twice will map a vector to itself, so $A^2 = I$.  This can be verified algebraically, as well.
A: Try $A=\begin{bmatrix}0&1\\1&0\end{bmatrix}$.
