description/proof of that norms on finite-dimensional complex vectors space are equivalent
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of norm on real or complex vectors space.
- The reader knows a definition of equivalence relation on norms on vectors space.
- The reader admits the proposition that any complex vectors space can be regarded to be the canonical real vectors space.
- The reader admits the proposition that any norms on any finite-dimensional real vectors space are equivalent.
Target Context
- The reader will have a description and a proof of the proposition that any norms on any finite-dimensional complex vectors space are equivalent.
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:
\(V\): \(\in \{\text{ the finite-dimensional complex vectors spaces }\}\)
\(S\): \(= \{\text{ the norms on } V\}\)
\(\sim\): \(\subseteq S \times S\), \(= \text{ the equivalence relation on } S\)
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Statements:
\(\forall s_1, s_2 \in S (s_1 \sim s_2)\)
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2: Proof
Whole Strategy: Step 1: take any basis for \(V\) and take the canonical bijection, \(g: V \to \mathbb{R}^{2 d}\); Step 2: see that for each norm on \(V\), the canonical norm is induced on \(\mathbb{R}^{2 d}\); Step 3: see that any 2 induced norms on \(\mathbb{R}^{2 d}\) are equivalent; Step 4: conclude the proposition.
Step 1:
Let us take any basis for \(V\), \(\{b_1, ..., b_d\}\).
There is the canonical bijection with respect to the basis, \(g: V \to \mathbb{R}^{2 d}: V \to \mathbb{C}^d \to \mathbb{R}^{2 d}, v = v^j b_j \mapsto (v^1 = w^1 + x^1 i, ..., v^d = w^d + x^d i) \mapsto (w^1, x^1, ..., w^d, x^d)\).
Let us regard \(\mathbb{R}^{2 d}\) as the Euclidean vectors space.
Let us see that \(g^{-1}: \mathbb{R}^{2 d} \to V\) is \(\mathbb{R}\)-linear: \(V\) can be regarded to be a \(\mathbb{R}\)-'vectors space', by the proposition that any complex vectors space can be regarded to be the canonical real vectors space.
For each \(r_1, r_2 \in \mathbb{R}\) and each \(u_1, u_2 \in \mathbb{R}^{2 d}\), \(g^{-1} (r_1 u_1 + r_2 u_2) = g^{-1} (r_1 (w_1^1, x_1^1, ..., w_1^d, x_1^d) + r_2 (w_2^1, x_2^1, ..., w_2^d, x_2^d)) = g^{-1} (r_1 w_1^1 + r_2 w_2^1, r_1 x_1^1 + r_2 x_2^1, ..., r_1 w_1^d + r_2 w_2^d, r_1 x_1^d + r_2 x_2^d) = ((r_1 w_1^1 + r_2 w_2^1) + (r_1 x_1^1 + r_2 x_2^1) i) b_1 + ... + ((r_1 w_1^d + r_2 w_2^d) + (r_1 x_1^d + r_2 x_2^d) i) b_d = (r_1 w_1^1 + r_1 x_1^1 i) b_1 + ... + (r_1 w_1^d + r_1 x_1^d i) b_d + (r_2 w_2^1 + r_2 x_2^1 i) b_1 + ... + (r_2 w_2^d + r_2 x_2^d i) b_d = r_1 ((w_1^1 + x_1^1 i) b_1 + ... + (w_1^d + x_1^d i) b_d) + r_2 ((w_2^1 + x_2^1 i) b_1 + ... + (w_2^d + x_2^d i) b_d) = r_1 g^{-1} (w_1^1, x_1^1, ..., w_1^d, x_1^d) + r_2 g^{-1} (w_2^1, x_2^1, ..., w_2^d , x_2^d) = r_1 g^{1} (u_1) + r_2 g^{1} (u_2)\).
Step 2:
Let us take any \(s_k \in S\).
Let us take the norm on \(\mathbb{R}^{2 d}\) induced by \(s_k\) under \(g\), \(t_k: \mathbb{R}^{2 d} \to \mathbb{R}, u \mapsto s_k (g^{-1} (u))\).
Let us see that \(t_k\) is indeed a norm: \(\mathbb{R}\)-linearity of \(g^{-1}\) is used.
1) \((0 \le t_k (u_1)) \land ((0 = t_k (u_1)) \iff (u_1 = 0))\): \(0 \le s_k (g^{-1} (u_1)) = t_k (u_1)\); if \(u_1 = 0\), \(g^{-1} (u_1) = 0\), so, \(t_k (u_1) = s_k (g^{-1} (u_1)) = 0\), and if \(t_k (u_1) = s_k (g^{-1} (u_1)) = 0\), \(g^{-1} (u_1) = 0\), so, \(u_1 = 0\).
2) \(t_k (r u_1) = \vert r \vert t_k (u_1)\): \(t_k (r u_1) = s_k (g^{-1} (r u_1)) = s_k (r g^{-1} (u_1)) = \vert r \vert s_k (g^{-1} (u_1)) = \vert r \vert t_k (u_1)\).
3) \(t_k (u_1 + u_2) \le t_k (u_1) + t_k (u_2)\): \(t_k (u_1 + u_2) = s_k (g^{-1} (u_1 + u_2)) = s_k (g^{-1} (u_1) + g^{-1} (u_2)) \le s_k (g^{-1} (u_1)) + s_k (g^{-1} (u_2)) = t_k (u_1) + t_k (u_2)\).
So, \(t_k\) is a norm on \(\mathbb{R}^{2 d}\).
Step 3:
For each \(s_j, s_l \in S\), \(t_j\) and \(t_l\) are equivalent, by the proposition that any norms on any finite-dimensional real vectors space are equivalent.
Step 4:
So, there are some \(c_1, c_2 \in \mathbb{R}\) such that \(0 \lt c_1, c_2\) such that for each \(v \in V\), \(c_1 t_l (g (v)) \le t_j (g (v)) \le c_2 t_l (g (v))\).
That means that \(c_1 s_l (g^{-1} (g (v))) \le s_j (g^{-1} (g (v))) \le c_2 s_l (g^{-1} (g (v)))\), which means that \(c_1 s_l (v) \le s_j (v) \le c_2 s_l (v)\).
That means that \(s_j\) and \(s_l\) are equivalent.
