 Solution

Put $P=Q{Q}^T$. Since $Q$ has $r$ columns, the rank bound for a matrix product gives $\mathrm{rank}(PA)\le r$. The Eckart-Young-Mirsky theorem and part (a) then giveOne can also see the lower bound directly. The restriction of $PA$ to the $(r+1)$-dimensional linear span of $v_1,\dots ,v_{r+1}$ has a nonzero null space, because its matrix rank is at most $r$. For a unit vector $z$ in that null space, $\Vert (A-PA)z{\Vert }_2=\Vert Az{\Vert }_2\ge {\sigma }_{r+1}$ by the singular value decomposition.