How to interpret involutory change of basis transformation? Just working through an assignment and a change of basis matrix popped up which was involutory - its own inverse.
I am not quite sure how to think about this... Presumably it means that the transformation doesn't 'scale' the basis vectors from one basis to another - this makes sense in the finite dimensional case since the determinant of any matrix representing the transformation must be $\pm 1$.
It was a 3-dimensional case, so is there some geometric interpretation?
What about a more general case? Arbitrary finite or infinite-dimensional?
 A: Obviously, to be involutory (at list in the finite dimension case), it has to be diagonalizable, and -- as you have noticed -- its eigenvalues must be $\pm 1$. In other words, your transformation $T$ has to have a form
$$T = S \Lambda S^{-1}, \quad \Lambda = \operatorname{diag}(\lambda_1,\dots,\lambda_n), \quad \lambda_k \in \{-1, 1\}.$$
The interpretation would be that your operator is a reflector along some of the basis vectors in some (not necessarily orthogonal!) basis, defined by the columns of $S$.
A: Even in the infinite-dimensional case, assuming the Axiom of Choice, there will be a basis for the vector space that consist of $T$-eigenvectors with eigenvalues $\pm 1$.
Well-order the vector space and process all vectors in sequence. For each nonzero vector $v$ we will add zero, one, or two new basis vectors, such that $v$ is in the span of the partial basis so far.
Consider the vectors $v+Tv$ and $v-Tv$. If one of them is zero, then $v$ is an eigenvector with eigenvalue $\pm 1$. Add it to the basis if it isn't already a combination of the basis vectors we've chosen so far.
If $v\pm Tv$ are both nonzero, then they are eigenvectors with eigenvalues $\pm 1$ (since $T(v\pm Tv)=Tv\pm T^2v = Tv\pm v$). Add each of them to the basis if they aren't already generated by it. Since $v=\frac 12(v+Tv)+\frac 12(v-Tv)$, it is now in the span of the basis so far.
