# prove that $A^2 = I$ implies that $A=I$ or $A= -I$ [duplicate]

How do we prove the above proposition?
$(AA)(ij)=1$ for $i=j$ , $0$ for $i\ne j$.
So, $[AA]_{ij} = \sum_k [A]_{ik} [A]_{kj}$.
How can I proceed from this point?

• What about $$\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}$$? Mar 8, 2016 at 21:45
• Over an arbitrary field $\mathbb{K}$ of characteristic not equal to $2$, there are exactly $n+1$ matrices, up to conjugacy, of dimension $n\times n$ which satisfy the condition. If $\text{char}\,\mathbb{K}=2$, then there are exactly $\left\lfloor \frac{n}{2}\right\rfloor+1$ matrices, up to conjugacy, of dimension $n\times n$ satisfying the condition. Mar 8, 2016 at 22:08
• Probably a duplicate of math.stackexchange.com/questions/106070/… and math.stackexchange.com/questions/44341/… Mar 16, 2016 at 5:56

Suppose $A$ is diagonal matrix then $A^2$ is diagonal with diagonal elements squares of diagonal elements of $A$...

Consider

$A=\begin{bmatrix} 1&0\\0&-1\end{bmatrix}$

$A=\begin{bmatrix} 1&0&0\\0&-1&0\\0&0&1\end{bmatrix}$

and many more..

This is not necessarily true. You should attempt to find a counter example in 2x2 matrices.

This is not generally true. It only implies that $A = A^{-1}$, i.e., that $A$ is its own inverse.

• Do you have a counterexample? Mar 16, 2016 at 5:20
• @MichaelHardy $\left[\begin{array}{ccc}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right]$ Mar 16, 2016 at 5:35