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.