Composition of Linear Isometries is Linear Isometry

Theorem

Let $\GF \in \set {\R, \C}$.

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

Let $T : X \to Y$ and $S : Y \to Z$ be linear isometries.

Then $S T$ is a linear isometry.

Proof

From Composition of Linear Transformations is Linear Transformation, $S T$ is a linear transformation.

For $x \in X$, we have:

\(\ds \norm {S T x}_Z\)

\(=\)

\(\ds \norm {T x}_Y\)

$S$ is a linear isometry

\(\ds \)

\(=\)

\(\ds \norm x_X\)

$T$ is a linear isometry

So $S T$ is a linear isometry.

$\blacksquare$