description/proof of that for real or complex vectors space, taking basis and sum of absolute components for each vector is norm
Topics
About: vectors space
The table of contents of this article
Starting Context
Target Context
- The reader will have a description and a proof of the proposition that for any real or complex vectors space, taking any basis and the sum of the absolute components for each vector is a norm.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(F\): \(\in \{\mathbb{R}, \mathbb{C}\}\), with the canonical field structure
\(V\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(B\): \(= \{b_j \vert j \in J\}\), \(\in \{\text{ the bases for } V\}\)
\(\Vert \bullet \Vert\): \(: V \to \mathbb{R}, v = v^{j_1} b_{j_1} + ... + v^{j_l} b_{j_l} \mapsto \vert v^{j_1} \vert + ... + \vert v^{j_l} \vert\)
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Statements:
\(\Vert \bullet \Vert \in \{\text{ the norms on } V\}\)
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2: Proof
Whole Strategy: Step 1: see that \(\Vert \bullet \Vert\) is well-defined; Step 2: see that \(\Vert \bullet \Vert\) satisfies the conditions to be a norm.
Step 1:
Let us see that \(\Vert \bullet \Vert\) is well-defined.
For each \(v \in V\), the decomposition with nonzero components, \(v = v^{j_1} b_{j_1} + ... + v^{j_l} b_{j_l}\), is unique (when \(v = 0\), it is \(v = 0\)), by the proposition that for any module with any basis, the components set of any element with respect to the basis is unique.
If \(v = v^{j_1} b_{j_1} + ... + v^{j_l} b_{j_l}\) has some zero components terms, the zero components terms do not influence the result.
When \(v = 0\), the components are inevitably zero, so, \(\Vert v \Vert = 0\).
When \(v \neq 0\), \(v = v^{j_1} b_{j_1} + ... + v^{j_l} b_{j_l}\) is the unique decomposition with nonzero components plus some zero components terms, and \(\vert v^{j_1} \vert + ... + \vert v^{j_l} \vert\) is uniquely determined.
\(\vert v^{j_1} \vert + ... + \vert v^{j_l} \vert \in \mathbb{R}\).
So, \(\Vert \bullet \Vert\) is well-defined.
Step 2:
Let us see that \(\Vert \bullet \Vert\) satisfies the conditions to be a norm.
Let \(v_1, v_2 \in V\) and \(r \in F\) be any.
1) \((0 \le \Vert v_1 \Vert) \land ((0 = \Vert v_1 \Vert) \iff (v_1 = 0))\): \(\Vert v_1 \Vert = \Vert v_1^{j_1} b_{j_1} + ... + v_1^{j_l} b_{j_l} \Vert = \vert v_1^{j_1} \vert + ... + \vert v_1^{j_l} \vert\), so, \(0 \le \Vert v_1 \Vert\); when \(v_1 = 0\), \(\Vert v_1 \Vert = 0\) as is seen in Step 1, while when \(\Vert v_1 \Vert = 0\), if \(v_1 \neq 0\), \(v_1 = v^{j_1} b_{j_1} + ... + v^{j_l} b_{j_l}\) with some nonzero components, and \(0 \lt \vert v^{j_1} \vert + ... + \vert v^{j_l} \vert\), a contradiction, so, \(v_1 = 0\).
2) \(\Vert r v_1 \Vert = \vert r \vert \Vert v_1 \Vert\): \(\Vert r v_1 \Vert = \Vert r (v_1^{j_1} b_{j_1} + ... + v_1^{j_l} b_{j_l}) \Vert = \vert r v_1^{j_1} \vert + ... + \vert r v_1^{j_l} \vert = \vert r \vert \vert v_1^{j_1} \vert + ... + \vert r \vert \vert v_1^{j_l} \vert = \vert r \vert (\vert v_1^{j_1} \vert + ... + \vert v_1^{j_l} \vert) = \vert r \vert \Vert v_1 \Vert\).
3) \(\Vert v_1 + v_2 \Vert \le \Vert v_1 \Vert + \Vert v_2 \Vert\): \(\Vert v_1 + v_2 \Vert = \Vert v_1^{m_1} b_{m_1} + ... + v_1^{m_n} b_{m_n} + v_2^{o_1} b_{o_1} + ... + v_2^{o_p} b_{o_p} \Vert\); now, denote \(\{b_{m_1}, ..., b_{m_n}\} = \{b_m \in B \vert m \in M\}\) and \(\{b_{o_1}, ..., b_{o_p}\} = \{b_o \in B \vert o \in O\}\), then, \(M \cap O\) may not be empty, and \(v_1^{m_1} b_{m_1} + ... + v_1^{m_n} b_{m_n} = \sum_{m \in M} v_1^m b_m = \sum_{m \in M \setminus O} v_1^m b_m + \sum_{m \in M \cap O} v_1^m b_m\) and \(v_2^{o_1} b_{o_1} + ... + v_2^{o_p} b_{o_p} = \sum_{o \in O} v_2^o b_o = \sum_{o \in O \setminus M} v_2^o b_o + \sum_{o \in O \cap M} v_2^o b_o\), so, \(v_1^{m_1} b_{m_1} + ... + v_1^{m_n} b_{m_n} + v_2^{o_1} b_{o_1} + ... + v_2^{o_p} b_{o_p} = \sum_{m \in M \setminus O} v_1^m b_m + \sum_{m \in M \cap O} v_1^m b_m + \sum_{o \in O \setminus M} v_2^o b_o + \sum_{o \in O \cap M} v_2^o b_o = \sum_{m \in M \setminus O} v_1^m b_m + \sum_{o \in O \setminus M} v_2^o b_o + \sum_{m \in M \cap O} (v_1^m + v_2^m) b_m\), and \(\Vert v_1 + v_2 \Vert = \sum_{m \in M \setminus O} \vert v_1^m \vert + \sum_{o \in O \setminus M} \vert v_2^o \vert + \sum_{m \in M \cap O} \vert v_1^m + v_2^m \vert \le \sum_{m \in M \setminus O} \vert v_1^m \vert + \sum_{o \in O \setminus M} \vert v_2^o \vert + \sum_{m \in M \cap O} (\vert v_1^m \vert + \vert v_2^m \vert) = \sum_{m \in M \setminus O} \vert v_1^m \vert + \sum_{m \in M \cap O} \vert v_1^m \vert + \sum_{o \in O \setminus M} \vert v_2^o \vert + \sum_{o \in M \cap O} \vert v_2^o \vert = \sum_{m \in M} \vert v_1^m \vert + \sum_{o \in O} \vert v_2^o \vert = \Vert v_1 \Vert + \Vert v_2 \Vert\).
3: Note
So, while we have proved the proposition that for any finite-dimensional real or complex vectors space, taking any basis and the sum of the absolute components for each vector is a norm, the finite-dimensional requirement is really unnecessary.
