= Frobenius rank inequality
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{title2=$\operatorname{rank}(PQ)+\operatorname{rank}(QR)\leq\operatorname{rank}(Q)+\operatorname{rank}(PQR)$}
For compatible <matrices> $P,Q,R$, the inequality is $\operatorname{rank}(PQ)+\operatorname{rank}(QR)\leq\operatorname{rank}(Q)+\operatorname{rank}(PQR)$. One proof uses the block <matrix> $M=\begin{pmatrix}PQ&0\\Q&QR\end{pmatrix}$. Subtract $P$ times its lower block row from its upper block row and then subtract its first block column times $R$ from its second. The result is $\begin{pmatrix}0&-PQR\\Q&0\end{pmatrix}$, with rank $\operatorname{rank}(Q)+\operatorname{rank}(PQR)$. On the other hand, choose independent columns of $PQ$ from the first block column and independent columns of $QR$ from the second. The corresponding columns of $M$ are independent: project a relation onto the upper block first, and then onto the lower block. Thus $\operatorname{rank}(M)\geq\operatorname{rank}(PQ)+\operatorname{rank}(QR)$.
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