Restriction of Norm on Vector Space to Subspace is Norm
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Theorem
Let $\Bbb F$ be a subfield of $\C$.
Let $X$ be a vector space over $\Bbb F$.
Let $\norm \cdot_X : X \to \hointr 0 \infty$ be a norm on $X$.
Let $Y$ be a vector subspace of $Y$.
Then $\norm \cdot_Y$, the restriction of $\norm \cdot_X$ to $Y$ is a norm on $Y$.
Proof
Since $\norm x_X \ge 0$ for any $x \in X$, we have $\norm y_Y \ge 0$ for any $y \in Y$.
We verify each of the axioms for a norm on a vector space.
Proof of $(\text N 1)$
Note that for $y \in Y$ we have $\norm y_Y = 0$ if and only if $\norm y_X = 0$.
Then by $(\text N 1)$ for $\norm \cdot_X$, we have $y = 0$.
$\Box$
Proof of $(\text N 2)$
Let $\lambda \in \Bbb F$ and $y \in Y$.
Then, we have:
\(\ds \norm {\lambda y}_Y\) | \(=\) | \(\ds \norm {\lambda y}_X\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \cmod \lambda \norm y_X\) | from $(\text N 2)$ for $\norm \cdot_X$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \cmod \lambda \norm y_Y\) |
$\Box$
Proof of $(\text N 3)$
Let $x, y \in Y$.
Then, we have:
\(\ds \norm {x + y}_Y\) | \(=\) | \(\ds \norm {x + y}_X\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \norm x_X + \norm y_Y\) | from $(\text N 3)$ for $\norm \cdot_X$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \norm x_Y + \norm y_Y\) |
$\blacksquare$