Let$A$ be a $3\times3$ real symmetric matrix such that $A^6=I$ . Then $A^2=I$

Let$A$ be a $3\times3$ real symmetric matrix such $ A^6=I$ . Then $A^2=I$.

How can I prove this statement is true or false?

As it is given $A$ is symmetric so $A=A^T$. Also $ A^6=I$.

But the main problem is that I can't operate $A^{-1}$ on both sides whether it is invertible or not. Can any help me what should I do?

Solution 1:

Since $A$ is real and symmetric, it is diagonalizable. So we may assume that $A$ is diagonal. In that case, we have $a_{jj}^6=1$ for all $j$. So $a_{jj}^2$ is nonnegative and its cube is $1$: thus $a_{jj}^2=1$, and $A^2=I$.

Note that this works for any diagonalizable $A$ (within $M_n(\mathbb R))$; it doesn't have to be symmetric.

Solution 2:

$A^6=I$ gives $x^6-1$ $=(x^2-1)(x^2-x+1)(x^2+x+1)$ is an annihilating polynomial of $A$. Let $m(x)$ be the minimal polynomial of $A$ then $m(x)$ divides $x^6-1$.
Since all eigenvalues of a real symmetric matrix are real and $x^2\pm x+1$ does not contain any real root, $m(x)$ cannot contain any factor of $x^2-x+1$ or $x^2+x+1$. Thus $m(x)$ divides $x^2-1$, which gives $x^2-1$ is an annihilating polynomial of $A$.