Continuous Linear Operator over Finite Dimensional Vector Space is Invertible

Theorem
Let $\struct {X, \norm {\, \cdot\,}_X}$ be a normed vector space.

Let $\map {CL} X := \map {CL} {X, X}$ be a continuous linear transformation space.

Let $I \in \map {CL} X$ be the identity element.

Let $S, T \in \map {CL} X$.

Suppose the dimension of $X$ is finite:


 * $d = \dim X < \infty$

Then $T \circ S = S \circ T = I$

Proof
Let $x \in X$.

Then:

Let $\mathbf 0 \in X$ be the zero vector of $X$.

Suppose $\map S x = \mathbf 0$.

Then:

Hence:


 * $\paren {\map S x = \mathbf 0} \implies \paren{x = \mathbf 0} \paren \star$

In other words:


 * $\map \ker S = \set {\mathbf 0}$

where $\ker$ denotes the kernel.

Let $\set {v_1, \ldots v_d}$ be the basis for $X$.

Suppose $\forall k \in \N_{> 0} : k \le d : \alpha_k \in \R$ are scalars such that:


 * $\ds \sum_{k \mathop = 1}^d \alpha_k \map S {v_k} = \mathbf 0$

By definition of linear transformation on vector space:


 * $\ds \map S {\sum_{k \mathop = 1}^d \alpha_k v_k} = \mathbf 0$

By Eq. $\paren \star$:


 * $\ds \sum_{k \mathop = 1}^d \alpha_k v_k = \mathbf 0$

Since $\set {v_1, \ldots v_d}$ was arbitrary, we have that:


 * $\forall k \in \N_{> 0} : k \le d : \alpha_k = 0$

By definition of linearly independent sequence of vectors, $\set {\map S {v_1}, \ldots \map S {v_d} }$ is a basis for $X$.

Therefore:


 * $\ds \forall x \in X : \exists \beta_k \in \R : x = \sum_{k \mathop = 1}^d \beta_k \map S {v_k} = \map S {\sum_{k \mathop = 1}^d \beta_k v_k}$

Thus:

All in all:


 * $\paren {T \circ S = I} \implies \paren {T \circ S = S \circ T = I}$