Continuous Linear Operator over Finite Dimensional Vector Space is Invertible

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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:

$d = \dim X < \infty$

Suppose $T \circ S = I$ where $\circ$ denotes the composition of mappings.


Then $T$ and $S$ are invertible.


Proof

Let $x \in X$.

Then:

\(\ds \map {\paren {T \circ S} } x\) \(=\) \(\ds \map I x\)
\(\ds \leadsto \ \ \) \(\ds \map T {\map {S} x}\) \(=\) \(\ds x\)

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

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

Then:

\(\ds \map T {\map S x}\) \(=\) \(\ds \map T {\mathbf 0}\)
\(\ds \) \(=\) \(\ds \mathbf 0\) Linear Transformation Maps Zero Vector to Zero Vector
\(\ds \) \(=\) \(\ds x\)

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

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

where $\mathbb F$ is a number field.

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:

$\ds \forall x \in X : \forall k \in \N_{> 0} : k \le d : \exists \beta_k \in \mathbb F : x = \sum_{k \mathop = 1}^d \beta_k \map S {v_k}$

Thus:

\(\ds \forall x \in X: \, \) \(\ds \map S {\map T x}\) \(=\) \(\ds \map S {\map T {\sum_{k \mathop = 1}^d \beta_k \map S {v_k} } }\)
\(\ds \) \(=\) \(\ds \map S {\map T {\map S {\sum_{k \mathop = 1}^d \beta_k v_k} } }\) Definition of Linear Transformation on Vector Space
\(\ds \) \(=\) \(\ds \map S {\map I {\sum_{k \mathop = 1}^d \beta_k v_k} }\) $\map {\paren {T \circ S} } x = \map I x$
\(\ds \) \(=\) \(\ds \map S {\sum_{k \mathop = 1}^d \beta_k v_k}\) Definition of Identity Element
\(\ds \) \(=\) \(\ds x\)

All in all:

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

$\blacksquare$


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