= Solution
The <orthogonal projection> $P=QQ^T$ gives
$$
QM-A=Q(M-Q^TA)-(I_m-P)A.
$$
The 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 $Q^TQ=I_r$, gives the stronger exact formula
$$
\|QM-A\|_F^2=\|M-Q^TA\|_F^2+\|(I_m-P)A\|_F^2.
$$
Consequently
$$
\boxed{\|QQ^TA-A\|_F\leq\|QM-A\|_F},
$$
with equality exactly when $M=Q^TA$. Thus solving the <linear least-squares problem> for each column chooses the best coefficients in the prescribed <column space>.
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