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)?
 A: The following matrix is a conterexample $
A =
\left( {\begin{array}{cc}
 -1 & 0  \\
 0 & 1  \\
 \end{array} } \right)
$
A: "Most" (read: diagonalizable) matrices can be viewed simply as a list of numbers -- its eigenvalues -- in the right basis. When doing arithmetic with just this matrix (or with other matrices that diagonalize in the same basis), you just do arithmetic on the eigenvalues.
So, to find diagonalizable solutions to $A^2 = I$, we just need to write down a matrix whose eigenvalues satisfy $\lambda^2 = 1$ -- and any such matrix will do.
When thinking about matrices in this way -- as a list of independent numbers -- it makes it easy to think your way through problems like this.
A: A simple counterexample is $$A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} $$ We have $A \neq \pm I$, but $A^{2} = I$. 
A: In dimension $\geq 2$ take the matrix that exchanges two basis vectors ("a transposition")
A: 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 $\vec e^2=\pm1$.
