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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?

Martin Sleziak
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Mus
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    Try and find a matrix that isn't the identity matrix whose square is the identity. – IntegrateThis Feb 22 '18 at 03:53
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    Try $$\begin{bmatrix}0&1\1&0\end{bmatrix}^2$$ – robjohn Feb 22 '18 at 03:54
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    You are saying that if $A=\pm I,$ the $A^2=I.$ No one disputes that, but the question is asking whether there is any other example where $A^2=I.$ – saulspatz Feb 22 '18 at 03:55
  • After "Question" there isn't an equation in the question. Maybe "if $A^2=I$ there? – coffeemath Feb 22 '18 at 03:57
  • What other number multiplied by itself could give one? arent these two the only cases? – Mus Feb 22 '18 at 04:09
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    "What other number multiplied by itself could give one?" There aren't any other numbers (real or complex) with that property, however we aren't talking about numbers. We are talking about matrices, and there are matrices whose square is the identity matrix other than the identity matrix and it's additive inverse. – JMoravitz Feb 22 '18 at 04:18
  • You can also just try purely diagonal matrices. – Michael Feb 22 '18 at 04:27
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    Have you studied how to prove or disprove a statement? – edm Feb 22 '18 at 05:39

4 Answers4

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counterexample:

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

anon
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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.

Siong Thye Goh
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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 $$ \begin{pmatrix} \cos \theta & \sin \theta \\ \sin \theta & - \cos \theta \end{pmatrix} $$

David Holden
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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 $$\begin{bmatrix}\cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{bmatrix}.$$

amd
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