A Linear Map from a vector space to itself is called an operator. \mathcal{L}(V) = \mathcal{L}(V,V), which is the set of all operators on V. constituents a vector space V a Linear Map T \in \mathcal{L}(V,V) requirements T is, by the constraints above, an operator additional information injectivity is surjectivity in finite-dimensional operators Suppose V is finite-dimensional and T \in \mathcal{L}(V), then, the following statements are equivalent: T is invertable T is injective T is surjective THIS IS NOT TRUE IN infinite-demensional vector space OPERATORS! (for instance, backwards shift in \mathbb{F}^{\infty} is surjective but not injective.) Proof: From the above, 1 \implies 2 by definition of invertability. Then, we have that T is invertable. We desire that T is surjective. Given invertability, we have that \null T = \{0\}. By the rank-nullity theorem, we have that: \dim V = \dim range\ T + \dim null\ T = \dim range\ T +0= \dim range\ T. Now, given T is an operator, we have that range\ T \subset V. Attempting to extend a basis of range\ T (which, given it is a subspace of V, is a linearly independent list in V) to a basis of V will be the trivial extension. So range\ T = V, which is also the codomain of T. This makes T surjective, as desired. So 2 \implies 3. Now, we have that T is surjective, we desire that T is invertable. We essentially reverse-engineer the step before. Given rank-nullity theorem, we have that: \dim V = \dim range\ T + \dim null\ T. Now, given T is surjective, \dim range\ T = \dim V. Therefore, we have that \dim V = \dim V + \dim null\ T \implies 0 = \dim null\ T. This makes the null space of T be \{0\}. This makes T injective. Having shown T to be both surjective and injective, T is invertable, as desired. So 3 \implies 1. Having shown a loop in the statements, all of them are equivalent. operators on complex vector spaces have an eigenvalue See operators on complex vector spaces have an eigenvalue