QR factorization of an orthogonal matrix
Solution 1:
$A = QR$
Since $Q$ is calculated by computing the orthogonal basis for $A$, $Q = A$ in the case that $A$ is already orthogonal.
To find $R$, we can rearrange the equality to be $Q^{-1}$$A = R$.
Recall $Q = A$, so we substitute to find that $A^{-1}$$A$ $=$ $R$ $=$ $I$
Since $A$ is orthogonal, its inverse is equal to its transpose and we have $A^{T}$$A$ $=$ $R$ $=$ $I$.
Thus, when $A$ is orthogonal, its $QR$ factorization is $A = IA$