description/proof of that for complex vectors space with basis, canonical real vectors space has basis that is union of basis for complex vectors space and its multiplied by \(i\)
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of basis of module.
- The reader admits the proposition that any complex vectors space can be regarded to be the canonical real vectors space.
Target Context
- The reader will have a description and a proof of the proposition that for any complex vectors space with any basis, the canonical real vectors space has the basis that is the union of the basis for the complex vectors space and its multiplied by \(i\).
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:
\(V\): \(\in \{\text{ the complex vectors spaces }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(B\): \(\in \{\text{ the bases for } V\}\), \(= \{b_j \vert j \in J\}\)
\(B'\): \(= B \cup \{i b_j \vert j \in J\}\)
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Statements:
\(B' \in \{\text{ the bases for } V \text{ as the canonical real vectors space }\}\)
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2: Proof
Whole Strategy: Step 1: see that \(B'\) is linearly independent; Step 2: see that \(B'\) spans \(V\).
Step 1:
Let us see that \(B'\) is linearly independent.
Let \(r_{j_1} b_{j_1} + ... + r_{j_m} b_{j_m} + r_{l_1} i b_{l_1} + ... + r_{l_n} i b_{l_n} = 0\) where \(r_{j_1}, ..., r_{j_m}, r_{l_1}, ..., r_{l_n} \in \mathbb{R}\).
Letting \(\{b_{k_1}, ..., b_{k_p}\} := \{b_{j_1}, ..., b_{j_m}\} \cup \{b_{l_1}, ..., b_{l_n}\}\), the left hand side of the above equation is \(c_{k_1} b_{k_1} + ... + c_{k_p} b_{k_p}\) where \(c_{k_1}, ..., c_{k_p} \in \mathbb{C}\).
As \(B\) is linearly independent, \(c_{k_1} = ... = c_{k_p} = 0\).
Each \(r_{j_q}\) appears as the real part of a \(c_{k_r}\), which is \(0\) as \(c_{k_r} = 0\).
Each \(r_{l_q}\) appears as the imaginary part of a \(c_{k_r}\), which is \(0\) as \(c_{k_r} = 0\).
So, \(r_{j_1} = ... = r_{j_m} = r_{l_1} = ... = r_{l_n} = 0\).
That means that \(B'\) is linearly independent.
Step 2:
Let us see that \(B'\) spans \(V\).
For each \(v \in V\), \(v = c_{k_1} b_{k_1} + ... + c_{k_p} b_{k_p}\), because \(B\) spans \(V\).
\(c_{k_q} = r_{k_q} + r'_{k_q} i\) where \(r_{k_q}, r'_{k_q} \in \mathbb{R}\).
So, \(v = (r_{k_1} + r'_{k_1} i) b_{k_1} + ... + (r_{k_p} + r'_{k_p} i) b_{k_p} = r_{k_1} b_{k_1} + ... + r_{k_p} b_{k_p} + r'_{k_1} i b_{k_1} + ... + r'_{k_p} i b_{k_p}\).
That means that \(B'\) spans \(V\).
