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If $A^2 = I$, then $A^n = I$ for all integers $n \geq 2$ (matrices)

Let $A=\pmatrix{1&0\\0&1}$

We see that $A^2$ = $\pmatrix{1&0\\0&1}\pmatrix{1&0\\0&1} = \pmatrix{1&0\\0&1}$

It seems that $A^n = I$ is true then, since I can just keep multiplying by the same matrix to get the identity matrix. My textbook however says False?

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    This ilmplies only $A^n=I$ for even $n$. – 2017-01-22
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    Maybe you mixed this up with: If $A^2=0$ then $A^n=0$ for all $n\ge 2$. – 2017-01-22
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    Try it for $A= -1$. – 2017-01-22
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    What do you mean $A=-1$? – 2017-01-22
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    $A = [-1]$ is a $1 \times 1$ matrix, whose only entry is $-1$. – 2017-01-22

4 Answers 4

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Let $A=\pmatrix{-1&0\\0&-1}$ then $A^2=I$

However $A^3=A$,

Therefore $A^n\neq I$ for $n=2k-1$, $k\in\mathbb Z^+$

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    How can you conclude that since $A^3$ doesn't work, no odd numbers work either? – 2017-01-22
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    You said If $A^2 = I$, then $A^n = I$ for all integers $n \geq 2$, and I have given you an inverse example which proves your equation is not true for some $A$ which is $A^2=I$. – 2017-01-22
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Really? If $A^2=I$, we must have $A^3=A$. Is your $A$ equal to $I$?

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    No but $A^2 = I$, and so when you say $A^3 = A$, that's saying $A^3 = I$ – 2017-01-22
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    So we have $A^3=A$ and $A^3=I$. Together that makes $A=I$. – 2017-01-22
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    I retract my argument – 2017-01-22
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    I have a question. When you say if $A^2 = I$, we must have $A^3= A$, how do you know the $A^3$ part? – 2017-01-22
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    @KSplitX If $A^2=I$, then $A(A^2)=A(I)$, i.e. $A^3=A$. – 2017-01-22
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Hint: Can you find an $A\neq I$ such that $A^2=I$? What is $A^3$?

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    Guess I never thought about it that way, but I can, and thus its False – 2017-01-22
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    It's true for _some_ $A$ (specifically, $A=I$), but not all, and therefore the statement is considered false. – 2017-01-22
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Search for Involutory Matrices.

If $$\begin{bmatrix} a & b \\ c & -a \end{bmatrix}$$ is such that $a^2+bc=1$, then $A^2=I$ but not neccesarily $A^3=I$. Take for example, $a=0$ and $b=c=1,$ then $$A^2=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\cdot \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=I$$ but $A^3=A \neq I.$