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:
- $d = \dim X < \infty$
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:
- $\paren {T \circ S = I} \implies \paren {T \circ S = S \circ T = I}$
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 2.4$: Composition of continuous linear transformations