Let $A$ symmetric matrix, and $A^2\ne I$.

Question

Let $A$ symmetric matrix, and $A^2\ne I$.

Prove there doesn't exist $k\in\mathbb{N}$ such that $A^k=I$

Any help would be appreciated.

Answer 1

Spectral Theorem says that $A$ is orthogonally diagonalizable, that is $\exists B$ orthogonal matrix and $D$ diagonal matrix such that $$ A=B^{-1}DB. $$ Now $$ A^k=(B^{-1}DB)(B^{-1}DB)\cdots (B^{-1}DB)=B^{-1}D^{k}B $$ so if by contradiction $A^k=I$ then we would have $$ B^{-1}D^kB=I $$ that is $$ D^k=I $$ that is $$ a_j^k=1 $$ where $a_j,\;j=1,\dots,n$ are the diagonal elements of $D$. This cannot happen if the entries are real and at least one among the $a_j$'s is different from $\pm1$, which follows by the hypotesis.

Answer 2

As stated the claim is false. Let $\xi$ be a $2n-$rooth of unite different to $\pm1$. Then the diagonal matrix $A$ whose diagonal entries are $\xi$ is symmetric and such that $A^2\neq I$, but $A^{2n}=I$. It is false also in the case that the entries $a_{ij}\in\Bbb F_q$ where $q=p^k$ for some prime $p$.

The claim is true if the matrix $A$ has real coefficient. Indeed $A$ is diagonalizable by the spectral theorem, that is there exists a matrix $C$ such that $$ A=CDC^{-1}$$ where $C$ is an orthogonal matrix and $D$ a diagonal matrix. Then $$A^{k}=CD^{k}C^{-1}=I$$ if and only if $d_{ii}^k=1$ over the reals, that is $d_{ii}=\pm1$, that is $D^2=I$, that is $A^2=I$.