If $A^2 = I$ (Identity Matrix) then $A = \pm I$ So I'm studying linear algebra and one of the self-study exercises has a set of true or false questions. One of the questions is this:

If $A^2 = I$ (Identity Matrix), then $A = \pm I$ ?

I'm pretty sure it is true but the answer says it's false. How can this be false (maybe it's a typography error in the book)?

I know $2·\mathbb C^2$ many counterexamples, namely
$$A=c_1\begin{pmatrix}
0&1\\
1&0
\end{pmatrix}+c_2\begin{pmatrix}
1&0\\
0&-1
\end{pmatrix}\pm\sqrt{c_1^2+c_2^2\pm1}\begin{pmatrix}
0&-1\\
1&0
\end{pmatrix},$$
see Pauli Matrices $\sigma_i$.
These are all such matrices and can be written as $A=\vec e· \vec \sigma$, where $\vec e^2=\pm1$.
