Why doesn't $A^2=I$ imply $A=\pm I$?

Question

Im having trouble believing this T/F Question: if $\mathrm A^2=I$ then $\mathrm A = \pm \mathrm I$

The answer is False but why?

If the matrix is $\mathrm A = \mathrm I,$ say

\begin{bmatrix}1& 0\\ 0 & 1 \end{bmatrix}

then $\mathrm A^2$ is also that. And if $\mathrm A = -\mathrm I,$ then it is

\begin{bmatrix}-1 & 0\\ 0 &-1 \end{bmatrix}

and that squared is also the same? Where am i going wrong?

Answer 1

counterexample:

$$A=\left(\begin{smallmatrix} 1 & 0\\ 0 & -1\end{smallmatrix}\right).$$

Answer 2

You were attemping to prove that if $A^2=I$, then $A=\pm I$ (it's a false statement).

Your attempt however was considering $A=\pm I$, then $A^2=I$, you were attempting to prove $D \implies C$ though the question was asking for $C \implies D$.

As shown by the other answer, $Q = \begin{bmatrix}1 & 0 \\ 0 & -1 \end{bmatrix}$ is a counter example.

Answer 3

for matrices in $\mathbb{M}_2(\mathbb{R})$ $$ \begin{pmatrix} a & b \\ c & d \end{pmatrix}^2 = \begin{pmatrix} a^2+bc & b(a+d) \\ c(a+d) & d^2+bc \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $$ requires either $b=c=0, a,d= \pm1$ or $a=-d, bc=1-a^2$

the first type of solution gives $4$ solutions which form a multiplicative group isomorphic to $V_4$ (Klein's Viergruppe).

the second type of solution gives a two-parameter family of matrices: $$ \begin{pmatrix} k & r \\ \frac{1-k^2}r & -k \end{pmatrix} $$ in case $|k| \le 1$ we may also (after re-scaling $r$) write these latter matrices as: $$ \begin{pmatrix} \cos \theta & r \sin \theta \\ \frac{\sin \theta}r & -\cos \theta \end{pmatrix} $$ looking at the subfamily with $r=1$ we obtain the solutions $$ \begin{pmatrix} \cos \theta & \sin \theta \\ \sin \theta & - \cos \theta \end{pmatrix} $$

Answer 4

It’s certainly true that if $A=\pm I$ then $A^2=I$, but that’s the converse of the statement in the question. There are plenty of other matrices that are their own inverse: for instance, $A^2=I$ is true of every reflection, so there’s an infinite number of counterexamples of the form $$\begin{bmatrix}\cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{bmatrix}.$$