# Square root-related calculations with matrices

If $\mathbf A$ is an $n \times n$ matrix such that $\mathbf A^6 = \mathbf I_n$ (the identity matrix), is it true that either $\mathbf A^3 = \mathbf I_n$ or $\mathbf A^3 = \mathbf -I_n$?

I'm struggling to solve this question.

I re-wrote $\mathbf A^3$ as $\mathbf B$, which gives $\mathbf B^2 = \mathbf I_n$. I do know that, if $\mathbf B^2 = \mathbf I_n$, then it is not necessarily true that $\mathbf B = \pm\mathbf I$ (I had a look at the following link, though I haven't fully understood the explanations given: If $A^2 = I$ (Identity Matrix) then $A = \pm I$).

Can I (and if so, how do I) apply this fact to the case where $\mathbf B$ is the cube of another matrix $\mathbf A$?

• In this case, $\mathbf A$ is necessarily real. – Caleb Owusu-Yianoma Feb 21 '15 at 0:40

The answer is no. $$A=\left(\begin{smallmatrix}0&1&0&0&0&0\\0&0&1&0&0&0\\0&0&0&1&0&0\\0&0&0&0&1&0\\0&0&0&0&0&1\\1&0&0&0&0&0\end{smallmatrix}\right)$$
The above is a permutation matrix. Each power of it is a permutation matrix, and its order is six, so the first power that equals the identity is $A^6$. In fact, a similar example can be constructed for $n=5$, using a different permutation of order $6$.
Let $A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}.$
The condition that $A^n=I$ is equivalent to saying that all the eigenvalues of $A$ are $n^{th}$ roots of $1$. So, in two dimensions, the only matrices squaring to $I$ but not equal to $I$ or $-I$ are those having eigenvalues of $1$ and $-1$ - hence a characteristic matrix of $x^2-1$. This means that $$\pmatrix{a&&b\\c&&d}$$ squares to $I$ and is not $I$ or $-I$ if and only if: $$(a-x)(d-x)-bc=x^2-1$$ meaning $$ad-bc=-1$$ $$a+d=0.$$ An instance of this is Git Gud's example, but there are infinitely many such examples - in particular, any matrix of the form $$\pmatrix{\lambda && \alpha \\ \frac{1-\lambda^2}{\alpha} &&-\lambda}$$ suffices for any chose of $\lambda$ and $\alpha$.