 Solution

The orthogonal projection $P=Q{Q}^T$ givesThe first summand has every column in the column space of $Q$, and the second has every column in its orthogonal complement. They therefore have zero Frobenius inner product. The Pythagorean identity, together with $Q^{T}Q=I_r$, gives the stronger exact formulaConsequentlywith equality exactly when $M=Q^{T}A$. Thus solving the linear least-squares problem for each column chooses the best coefficients in the prescribed column space.