Composition of Linear Isometries is Linear Isometry
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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.
$\blacksquare$