invertible operator
An operator T : V -> V is invertible if there is an operator S that undoes it from both sides: S T = T S = I. That S, when it exists, is unique and is written T^-1. Invertibility means T is a bijection — it loses no information (injective) and reaches everything (surjective) — so it is a perfect, reversible reshuffling of the space.
In finite dimensions a small miracle happens: the three conditions injective, surjective, and invertible all coincide. This is a direct corollary of rank-nullity. If T : V -> V is injective then ker T = {0}, so dim im T = dim V, so im T = V and T is surjective; and a map that is both is a bijection, hence invertible. Checking either one of injectivity or surjectivity alone certifies the whole package. (This collapse fails in infinite dimensions, where one-sided behavior can occur.)
Equivalent finite-dimensional tests pile up: T is invertible iff its matrix has nonzero determinant, iff 0 is not an eigenvalue, iff its columns are linearly independent, iff it has full rank, iff its matrix can be row-reduced to the identity. Each is the same fact wearing different clothes; pick whichever is cheapest to check.
The invertible operators on V form a group under composition, the general linear group GL(V). It is not a subspace — the sum of two invertible operators can be singular — but it is closed under composition and inverse, with the identity as unit. GL(V) is the symmetry group of a vector space and the stage on which similarity and conjugation play out.
In finite dimensions, checking injectivity alone already proves invertibility — rank-nullity does the rest.
The injective = surjective = invertible collapse is special to finite dimensions. On the space of infinite sequences the right shift is injective but not surjective, and the left shift is surjective but not injective — neither is invertible, and each is only a one-sided inverse of the other.