If is a pseudo-differential operator and is a parametrix, this gives a right inverse to, in other words a fundamental solution, provided that is " small enough " which in practice means that it should be a sufficiently good smoothing operator.
22.
They call an element " x " right quasiregular if there exists " y " such that x + y + xy = 0, which is equivalent to saying that 1 + " x " has a right inverse when the ring is unital.
23.
The Jacobson radical of a ring consists of elements that satisfy a stronger property than being merely a non-unit in some sense, a member of the Jacobson radical must not " act as a unit " in " any " canonical homomorphism to the zero of every " right division ring " ( each non-zero element of which has a right inverse ) internal to the ring in question.
24.
The mapping T is injective, but we don't know that it's " surjective ", so B ( or B's columns-we can deal with them one at a time ) might not be in the range of T . If B is in the range of T, then A', being a left inverse for A, will map B back to its preimage, namely A'B . In that case, we're all good, and even though A'may not be a right inverse for A, we know that AA'will be a square matrix that maps B to B . On the other hand, if B is not in the range of T, then there is no solution.
