Invertible Continuous Linear Operator is Bijective

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