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.
Best Answer
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, $A^2=I$.