Use of rank theorem

Question

Let $$|rank(P)-rank(Q)|^2=(rank(P)-rank(Q))^2=rank(P)^2-2(rank(P))(rank(Q))+rank(Q)^2$$ $$|rank(P)-rank(Q)|^2\le |rank(P)|^2+2|rank(P)||rank(Q)|+|rank(Q)|^2$$ $$|rank(P)-rank(Q)|^2\le |rank(P)|+|rank(Q)|$$ Since, $rank(P)\ge 0$ and $rank(Q)\ge 0$ $$|rank(P)-rank(Q)|^2\le rank(P)+rank(Q)$$ I'm stuck here and I'm sure if my proof thus far is correct since we know that $$rank(P+Q)\le rank(P)+rank(Q)$$

Answer 1

I would not say that your proof is on the right track. I would suggest that you instead use the fact (or try to prove yourself) that $$ \operatorname{rank}(P + Q) \leq \operatorname{rank}(P) + \operatorname{rank}(Q). $$ Now, if we set $P = A + B$ and $Q = -B$ and note that $\operatorname{rank}(B) = \operatorname{rank}(-B)$, we find that $$ \operatorname{rank}((A + B) - B) \leq \operatorname{rank}(A + B) + \operatorname{rank}(-B) \implies\\ \operatorname{rank}(A) \leq \operatorname{rank}(A + B) + \operatorname{rank}(B) \implies\\ \operatorname{rank}(A) - \operatorname{rank}(B) \leq \operatorname{rank}(A + B). $$ Now, if we do the same thing but reverse the roles of $A$ and $B$, we end up with $$ \operatorname{rank}(B) - \operatorname{rank}(B) \leq \operatorname{rank}(A + B). $$ Putting these two results together gives us the desired inequality $$ |\operatorname{rank}(A) - \operatorname{rank}(B)| \leq \operatorname{rank}(A + B). $$