 Solution

The thin QR decomposition gives $AX=QR$. Its columns already belong to the column space of $Q$, so the orthogonal projection $Q{Q}^T$ leaves them fixed:This identity uses only $AX=QR$ and $Q^{T}Q=I_r$; it remains valid if $R$ is singular, provided such a factorization with orthonormal $Q$ is used.