1077: Complex Conjugate of Complex Vectors Space

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definition of complex conjugate of complex vectors space

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Topics

About: vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note

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Starting Context

• The reader knows a definition of %field name% vectors space.

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Target Context

• The reader will have a definition of complex conjugate of complex vectors space.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$\mathbb{C}$: with the canonical field structure
$V$: $\in \{\text{ the }\mathbb{C}\text{ vectors spaces }\}$, with any addition, $+$, and any scalar multiplication, $\ast$
$\bar{V}$: $=V$, $\in \{\text{ the }\mathbb{C}\text{ vectors spaces }\}$, with the addition, $\oplus$, and the scalar multiplication, $\otimes$, specified below
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Conditions:
$\mathrm{\forall }v,v^{\prime }\in \bar{V}(v\oplus v^{\prime }=v+v^{\prime })$
$\wedge$
$\mathrm{\forall }v\in \bar{V},\mathrm{\forall }c\in \mathbb{C}(c\otimes v=\bar{c}\ast v)$
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2: Note

$\bar{V}=V$ means that they are the same sets-wise.

There is the bijective correspondence, $f:V\to \bar{V},v\mapsto v$, which is not any vectors space homomorphism: $f(c\ast v)=c\ast v=\bar{c}\otimes v=\bar{c}\otimes f(v)$, which is not linear.

$f(v)$ is called "conjugate of $v$".

When $V$ is $d$-dimensional and has a basis, $B=\{b_1,...,b_d\}$, $B$ is also a basis for $\bar{V}$: for $c^1\otimes b_1\oplus ...\oplus c^d\otimes b_d=0$, $\overline{c^1}\ast b_1+...+\overline{c^d}\ast b_d=0$, which implies that $\overline{c^1}=...=\overline{c^d}=0$, which implies that $c^1=...=c^d=0$, so, $B$ is linearly independent on $\bar{V}$; for each $v\in \bar{V}$, $v=v^1\ast b_1+...+v^d\ast b_d=\overline{v^1}\otimes b_1\oplus ...\oplus \overline{v^d}\otimes b_d$, so, $B$ spans $\bar{V}$.

The components set of any $v\in V$ with respect to $B$ is $(v^1,...,v^d{)}^t$, which means that $v=v^1\ast b_1+...+v^d\ast b_d$, which means that $v=\overline{v^1}\otimes b_1\oplus ...\oplus \overline{v^d}\otimes b_d$, so, the components set of $v\in \bar{V}$ with respect to $B$ is $(\overline{v^1},...,\overline{v^d})$, which is frequently denoted as the row vector, as a custom.

When $V$ has an inner product, $⟨\bullet ,\bullet {⟩}_V$, $⟨\bullet ,\bullet ⟩:\bar{V}\times V\to \mathbb{C},(v^{\prime },v)\mapsto ⟨v,f^{-1}(v^{\prime }){⟩}_V$ is often called "inner product", although the definition of inner product on real or complex vectors space is on a single vectors space.

For $v^{\prime }=v^{\prime j}\otimes b_j$ and $v=v^l\ast b_l$, $⟨v^{\prime },v⟩=⟨v^{\prime j}\otimes b_j,v^l\ast b_l⟩=⟨v^l\ast b_l,f^{-1}(v^{\prime j}\otimes b_j){⟩}_V=⟨v^l\ast b_l,\overline{v^{\prime j}}\ast b_j{⟩}_V=\overline{v^{\prime j}}{v}^l⟨b_l,b_j{⟩}_V$, and especially if the basis is orthonormal, $=\overline{v^{\prime j}}{v}^l{\delta }_{l,j}=\sum _j\overline{v^{\prime j}}{v}^j=(\overline{v^{\prime 1}},...,\overline{v^{\prime d}})(v^1,...,v^d{)}^t$.

Let us see that $\bar{V}$ is indeed a $\mathbb{C}$ vectors space.

1) for any elements, $v_1,v_2\in \bar{V}$, $v_1\oplus v_2\in \bar{V}$ (closed-ness under addition): $v_1\oplus v_2=v_1+v_2\in V=\bar{V}$.

2) for any elements, $v_1,v_2\in \bar{V}$, $v_1\oplus v_2=v_2\oplus v_1$ (commutativity of addition): $v_1\oplus v_2=v_1+v_2=v_2+v_1=v_2\oplus v_1$.

3) for any elements, $v_1,v_2,v_3\in \bar{V}$, $(v_1\oplus v_2)\oplus v_3=v_1\oplus (v_2\oplus v_3)$ (associativity of additions): $(v_1\oplus v_2)\oplus v_3=(v_1+v_2)+v_3=v_1+(v_2+v_3)=v_1\oplus (v_2\oplus v_3)$.

4) there is a 0 element, $0\in \bar{V}$, such that for any $v\in \bar{V}$, $v\oplus 0=v$ (existence of 0 vector): $0\in V=\bar{V}$, and $v\oplus 0=v+0=v$.

5) for any element, $v\in \bar{V}$, there is an inverse element, $v^{\prime }\in \bar{V}$, such that $v^{\prime }\oplus v=0$ (existence of inverse vector): $v^{\prime }:=-v\in V=\bar{V}$, and $v^{\prime }\oplus v=-v+v=0$.

6) for any element, $v\in \bar{V}$, and any scalar, $r\in \mathbb{C}$, $r\otimes v\in \bar{V}$ (closed-ness under scalar multiplication): $r\otimes v=\bar{r}\ast v\in V=\bar{V}$.

7) for any element, $v\in \bar{V}$, and any scalars, $r_1,r_2\in \mathbb{C}$, $(r_1+r_2)\otimes v=r_1\otimes v\oplus r_2\otimes v$ (scalar multiplication distributability for scalars addition): $(r_1+r_2)\otimes v=\overline{(r_1+r_2)}\ast v=(\overline{r_1}+\overline{r_2})\ast v=\overline{r_1}\ast v+\overline{r_2}\ast v=r_1\otimes v\oplus r_2\otimes v$.

8) for any elements, $v_1,v_2\in \bar{V}$, and any scalar, $r\in \mathbb{C}$, $r\otimes (v_1\oplus v_2)=r\otimes v_1\oplus r\otimes v_2$ (scalar multiplication distributability for vectors addition): $r\otimes (v_1\oplus v_2)=\bar{r}\ast (v_1+v_2)=\bar{r}\ast v_1+\bar{r}\ast v_2=r\otimes v_1\oplus r\otimes v_2$.

9) for any element, $v\in \bar{V}$, and any scalars, $r_1,r_2\in \mathbb{C}$, $(r_1{r}_2)\otimes v=r_1\otimes (r_2\otimes v)$ (associativity of scalar multiplications): $(r_1{r}_2)\otimes v=\overline{(r_1{r}_2)}\ast v=(\overline{r_1}\overline{r_2})\ast v=\overline{r_1}\ast (\overline{r_2}\ast v)=r_1\otimes (r_2\otimes v)$.

10) for any element, $v\in \bar{V}$, $1\otimes v=v$ (identity of 1 multiplication): $1\otimes v=\bar{1}\ast v=1\ast v=v$.

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References

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