Infinite-dimensional inner product spaces: if $A^k = I$ for self-adjoint $A$ and for integer $k > 0$, then $A^2 = I$ Exercise 5(d), Sec 80, Pg 162, PR Halmos's Finite-Dimensional Vector Spaces:
If $A^k = I$ where $A$ is a self-adjoint operator and $k > 0$ is some positive integer, show that $A^2 = I$. The underlying inner product space is not necessarily finite-dimensional.

I see that it is rather straightforward to establish the result in finite-dimensional spaces (over both real and complex fields), using the Spectral Theorem for self-adjoint operators. However, I'm finding it difficult to extend the result to infinite dimensions for $k \geq 3$. Towards showing $A^2 = I$, my (unsuccessful) attempts so far have been around establishing $\Vert A^2x-x\Vert = 0$. Would appreciate some help. Thanks for reading.
 A: Since $A$ is self-adjoint, all eigenvalues of $A$ are real. Also, $\lambda$ is an eigenvalue of $A$ implies $Av=\lambda v$ for some non-zero $v$, i.e. $v=A^kv=\lambda^kv$. So, $\lambda^k=1\implies \lambda=\pm 1$.
Now, $0=A^k-I=\prod_j(A -\xi_jI)$, where $\xi_j$ are $k$-th roots of unity. Since, $\pm 1$ are the only eigenvalues of $A$, we have $(A-\xi_j I)u\not=0$ for any non-zero vector $u$ if $\xi_j\not=\pm 1$. In other words, $\prod_{\xi_j\not=\pm 1}(A-\xi_jI)$ is injective. But, $\prod_j(A -\xi_jI)=0$, so $A^2=I$.
$\textbf{Edit:---}$ So, let $T:=(A^2-I)\prod_{\xi_j\not=\pm 1}(A-\xi_jI)$. Then, $T=0$. Note that either $A-I$ or $A+I$ or both are factors of $\prod_{j}(A-\xi_jI)=0$. Also, let $S=\prod_{\xi_j\not=\pm 1}(A-\xi_jI)$, then, $T=S(A^2-I)=(A^2-I)S$, and $S$ is injective. Therefore, for any $v\not=0$, we have $0=0(v)=T(v)=S\big((A^2-I)v\big)$. If, $(A^2-I)v=0$ we are done, but if $w:=(A^2-I)v\not=0$, then $S(w)=0$, contradicts the fact that $S$ is injective.
A: For any inner product space, complete or not, let $v$ be any vector and consider the finite-dimensional subspace generated from $v,Av,\ldots,A^{k-1}v$. Then this space is $A$-invariant, so $A$ restricted to it is a symmetric matrix. As mentioned in the question, it is straightforward to show that $A^2=I$ in this space, that is, $A^2v=v$. Since this is true of any vector, $A^2=I$.
A: The spectrum of $A$ must be a subset of $\{-1,1\}$ by the spectral mapping theorem and the fact that $A$ is self-adjoint. Hence $A$ is unitary and $A=A^*=A^{-1}$, so $A^2=I$.
A: Let $P=A^2=A^\ast A$ and $S=P^{k-1}+P^{k-2}+\cdots+P+I$. By the given conditions, $P$ is a positive operator, $S$ is strictly positive and $S(P-I)=P^k-I=A^{2k}-I=0$. Hence $\langle S(P-I)x,(P-I)x\rangle=0$ for every vector $x$. As $S$ is strictly positive, this implies that $(P-I)x=0$ for every $x$. Therefore $P=I$.
