Invertible Continuous Linear Operator is Bijective
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Theorem
Let $\struct {X, \norm {\, \cdot \,} }$ be the 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.
Suppose $A \in \map {CL} X$ is invertible.
Then $A$ is bijective.
Proof
$A$ is injective
Let $x, y \in X$ be such that $\map A x = \map A y$.
Then:
- $A^{-1} \circ \map A x = A^{-1} \circ \map A y$
where $A^{-1}$ is the inverse of $A$.
By definition:
- $A^{-1} \circ A = I$
Hence:
- $x = y$
By definition, $A$ is injective.
$\Box$
$A$ is surjective
Let $y \in X$
Then:
- $x := \map {A^{-1} } y \in X$
Moreover:
\(\ds \map A x\) | \(=\) | \(\ds A \circ \map {A^{-1} } y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map I y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds y\) |
Hence:
- $\forall y \in X : \exists x \in X : \map A x = y$
By definition, $A$ is surjective.
$\Box$
By definition, $A$ is bijective.
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 2.4$: Composition of continuous linear transformations