A First Course in Linear Algebra

The converse of the implication \(P\Rightarrow Q\) is the implication \(Q\Rightarrow P\text{.}\) There is no guarantee that the truth of these two statements are related. In particular, if an implication has been proven to be a theorem, then do not try to use its converse too, as if it were a theorem. Sometimes the converse is true (and we have an equivalence, see Proof Technique E). But more likely the converse is false, especially if it was not included in the statement of the original theorem.

For example, we have the theorem, “if a vehicle is a fire truck, then it is has big tires and has a siren.” The converse is false. The statement that “if a vehicle has big tires and a siren, then it is a fire truck” is false. A police vehicle for use on a sandy public beach would have big tires and a siren, yet is not equipped to fight fires.

We bring this up now, because Theorem CSRN has a tempting converse. Does this theorem say that if \(r\lt n\text{,}\) then the system is consistent? Definitely not, as Archetype E has \(r=3\lt 4=n\text{,}\)] yet is inconsistent. This example is then said to be a counterexample to the converse. Whenever you think a theorem that is an implication might actually be an equivalence, it is good to hunt around for a counterexample that shows the converse to be false (the Archetypes, Appendix A, can be a good hunting ground).