Inverse of Linear Isometry is Linear Isometry

Theorem

Let $\struct {X, \norm \cdot_X}$ and $\struct {Y, \norm \cdot_Y}$ be normed vector spaces.

Let $T : X \to Y$ be an invertible (in the sense of a mapping) linear isometry with inverse $T^{-1} : Y \to X$.

Then $T^{-1}$ is a linear isometry.

Proof

From Inverse of Linear Transformation is Linear Transformation, we have:

$T^{-1}$ is a linear transformation.

Since $T$ is a linear isometry, we have:

$\norm {T x}_Y = \norm x_X$

for each $x \in X$.

Note that for each $y \in Y$, we have $T^{-1} y \in X$.

We then have:

\(\ds \norm {T^{-1} y}_X\)

\(=\)

\(\ds \norm {\map T {T^{-1} y} }_Y\)

\(\ds \)

\(=\)

\(\ds \norm {\map {\paren {T \circ T^{-1} } } y}_Y\)

\(\ds \)

\(=\)

\(\ds \norm y_Y\)

Definition of Inverse Mapping

So $T$ is a linear transformation with:

$\norm {T^{-1} y}_X = \norm y_Y$

for each $y \in Y$.

So $T^{-1}$ is a linear isometry.

$\blacksquare$