Find a $QR$ factorization of a matrix $A$, given that $A$ is orthogonal.
So we know that the QR factorization means that for a given $m \times n $ matrix $A$ with linearly independent columns, there exists a factorization $QR$ such that $Q$ is an $m \times n$ matrix with orthonormal columns and $R$ is an invertible upper triangular matrix.
I don't really know what extra insight the fact that $A$ is orthogonal provides us here. We now know that $A$ is a square matrix, which means that $Q$ is also a square matrix and as such, $Q$ is also an orthogonal matrix.
But I don't see how this can give us a generalised $QR$ factorization in this case, can anyone provide a hint?
Best Answer
$A = QR$
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 $A = IA$