Question

Show that C^n is a complex vector space

Answer 1

We have to show that $\mathbb{C}^n$ is a complex vector space.

According to Wikipedia's definition of the vector space we can conclude that:

A vector space over a field $\mathbb{F}$ - is a set $V$ together with two operations:

• vector addition ( $+$ ) : $V + V \mapsto V$ - takes any two vectors $v$ and $w$ and assigns to them a third vector which is commonly written as $v + w$, and called the sum of these two vectors. ( The resultant vector is also an element of the set $V$ )
• scalar multiplication ( $\cdot$ ) : $\mathbb{F} \cdot V \mapsto V$ - takes any scalar $\alpha$ and any vector $\upsilon$ and gives another vector $\alpha \cdot v$. ( Similarly, the vector $\alpha \cdot v$ is an element of the set $V$ ) ( Scalar multiplication is a multiplication of a vector by a scalar )

that satisfy the eight axioms listed below ( considering that $u, v, w$ are arbitrary vectors in $V$, and $\alpha, \beta$ scalars in $\mathbb{F}$ ):

1. associativity of addition : $u + (v + w) = (u + v) + w$
2. commutativity of addition : $u + v = v + u$
3. identity element of addition ( there exists an element $0 \in V$, called the zero vector, such that $v + 0 = v$ for all $v \in V$ )
4. inverse element of addition ( for every $v \in V$, there exists an element $-v \in V$ , called the additive inverse of $v$ , such that $v + (-v) = 0$ )
5. associativity of multiplication : $\alpha \cdot (\beta \cdot v) = (\alpha \cdot \beta) \cdot v$
6. identity element of scalar multiplication : $1 \cdot v = v$ ( where $1$ denotes the multiplicative identity in $\mathbb{F}$ )
7. distributivity ( of scalar multiplication with respect to vector addition ) : $\alpha \cdot (u + v) = \alpha \cdot u + \alpha \cdot v$
8. distributivity ( of scalar multiplication with respect to field addition ) : $(\alpha + \beta) \cdot v = \alpha \cdot v + \beta \cdot v$

Elements of $V$ are commonly called vectors. Elements of $\mathbb{F}$ are commonly called scalars.

So, if we prove this 2 operation properties and 8 axioms for our $\mathbb{C}^n$ we will prove that it is a complex vector space.

VECTOR ADDITION

Let's consider $\mathbb{C}^1$ case:

Taking 2 elements from $\mathbb{C}^1$-space: $u = a+ib \in \mathbb{C}^1$ , and $v = c+id \in \mathbb{C}^1$ , we can write: $u + v = a+ib + c+id = (a+b) + i(c+d) = t \in \mathbb{C}^1.$

Now consider $\mathbb{C}^n$ case:

Taking 2 elements from $\mathbb{C}^n$-space: $u = (a_1+ib_1, a_2+ib_2, \ldots, a_n+ib_n) \in \mathbb{C}^n$ , and $v = (c_1+id_1, c_2+id_2, \ldots, c_n+id_n) \in \mathbb{C}^n$ , we can write: $u + v = (a_1+ib_1, a_2+ib_2, \ldots, a_n+ib_n) + (c_1+id_1, c_2+id_2, \ldots, c_n+id_n) = \newline ((a_1+b_1) + i(c_1+d_1), (a_2+b_2) + i(c_2+d_2), \ldots, (a_n+b_n) + i(c_n+d_n)) = t \in \mathbb{C}^n.$

$\square$

SCALAR MULTIPLICATION

The same way, let's consider $\mathbb{C}^1$ case:

Taking element from $\mathbb{C}^1$-space: $u = a+ib \in \mathbb{C}^1$ , and a scalar $\alpha \in \mathbb{R}$ , we can write: $\alpha \cdot u = \alpha \cdot (a+ib) = \alpha \cdot a + i \alpha \cdot b = t \in \mathbb{C}^1.$

Now consider $\mathbb{C}^n$ case:

Taking element from $\mathbb{C}^n$-space: $u = (a_1+ib_1, a_2+ib_2, \ldots, a_n+ib_n) \in \mathbb{C}^n$ , and a scalar $\alpha \in \mathbb{R}$ , we can write:

$\alpha \cdot u = \alpha \cdot (a_1+ib_1, a_2+ib_2, \ldots, a_n+ib_n) = \newline (\alpha \cdot (a_1+ib_1), \alpha \cdot (a_2+ib_2), \ldots, \alpha \cdot (a_n+ib_n) ) = \newline (\alpha \cdot a_1 + i \alpha \cdot b_1, \alpha \cdot a_2 + i \alpha \cdot b_2, \ldots, \alpha \cdot a_n + i \alpha \cdot b_n)= t \in \mathbb{C}^n.$

$\square$

1) ASSOCIATIVITY OF ADDITION

Let's take 3 elements $u, v, w$ from $\mathbb{C}^n$ -space: $u = (a_1+ib_1, \ldots, a_n+ib_n)$, $v = (c_1+id_1, \ldots, c_n+id_n)$, and $w = (f_1+ig_1, \ldots, f_n+ig_n)$, and show that $u + (v + w) = (u + v) + w.$

$u + (v + w) = \newline (a_1+ib_1, \ldots, a_n+ib_n) + \Big( (c_1+id_1, \ldots, c_n+id_n) + (f_1+ig_1, \ldots, f_n+ig_n) \Big) = (a_1+ib_1, \ldots, a_n+ib_n) + \Big( (c_1+f_1)+i(d_1+g_1), \ldots, (c_n+f_n)+i(d_n+g_n)\Big) = \newline \Big( (a_1+c_1+f_1)+i(b_1+d_1+g_1), \ldots, (a_n+c_n+f_n)+i(b_n+d_n+g_n)\Big) = \newline \Big( (a_1+c_1)+i(b_1+d_1), \ldots, (a_n+c_n)+i(b_n+d_n)\Big) + (f_1+ig_1, \ldots, f_n+ig_n) = \newline \Big( (a_1+ib_1, \ldots, a_n+ib_n) + (c_1+id_1, \ldots, c_n+id_n) \Big) + (f_1+ig_1, \ldots, f_n+ig_n) = \newline (u + v) + w.$

$\square$

2) COMMUTATIVITY OF ADDITION

Let's take 2 elements $u, v$ from $\mathbb{C}^n$-space: $u = (a_1+ib_1, \ldots, a_n+ib_n)$, and $v = (c_1+id_1, \ldots, c_n+id_n)$, and show that $u + v = v + u.$

$u + v = (a_1+ib_1, \ldots, a_n+ib_n) + (c_1+id_1, \ldots, c_n+id_n) = \newline \Big( (a_1+c_1)+i(b_1+d_1), \ldots, (a_n+c_n)+i(b_n+d_n)\Big) = \newline \Big( (c_1+a_1)+i(d_1+b_1), \ldots, (c_n+a_n)+i(d_n+b_n)\Big) = \newline (c_1+id_1, \ldots, c_n+id_n) + (a_1+ib_1, \ldots, a_n+ib_n) = v + u.$

$\square$

3) IDENTITY ELEMENT OF ADDITION

There exist $0 = (0_1, 0_2, \ldots, 0_n) \in \mathbb{C}^n$, such that using VECTOR ADDITION property (proven above) we can write $v + 0 = v, \forall v \in V.$

$\square$

4) INVERSE ELEMENT OF ADDITION

For every $v = (c_1+id_1, \ldots, c_n+id_n) \in V$, there exist element $-v = (-c_1-id_1, \ldots, -c_n-id_n)$, such that $v + (-v) = 0$ (according to VECTOR ADDITION property).

$\square$

5) ASSOCIATIVITY OF MULTIPLICATION

For any $v = (c_1+id_1, \ldots, c_n+id_n) \in V$, and 2 scalars $\alpha$ and $\beta$, let's prove that: $\alpha \cdot (\beta \cdot v) = (\alpha \cdot \beta) \cdot v$

$\alpha \cdot (\beta \cdot v) = \alpha \cdot \Big(\beta \cdot (c_1+id_1, \ldots, c_n+id_n)\Big) = \newline \alpha \cdot \Big(\beta \cdot c_1+i\beta \cdot d_1, \ldots, \beta \cdot c_n+i\beta \cdot d_n\Big) = \newline \Big(\alpha \cdot \beta \cdot c_1+i\alpha \cdot \beta \cdot d_1, \ldots, \alpha \cdot \beta \cdot c_n+i\alpha \cdot \beta \cdot d_n\Big) = \newline \Big( (\alpha \cdot \beta) \cdot c_1+(\alpha \cdot \beta) \cdot id_1, \ldots, (\alpha \cdot \beta) \cdot c_n+(\alpha \cdot \beta) \cdot id_n \Big) = \newline. (\alpha \cdot \beta) \cdot (c_1+id_1, \ldots, c_n+id_n) = (\alpha \cdot \beta) \cdot v$

$\square$

6) IDENTITY ELEMENT OF SCALAR MULTIPLICATION

There exist $1 = (1_1, 1_2, \ldots, 1_n) \in \mathbb{C}^n$, such that using VECTOR ADDITION property (proven above) we can write $1 \cdot v = v, \forall v \in V.$

$\square$

7) DISTRIBUTIVITY ( OF SCALAR MULTIPLICATION WITH RESPECT TO VECTOR ADDITION )

For any 2 elements $u, v$ from $\mathbb{C}^n$-space: $u = (a_1+ib_1, \ldots, a_n+ib_n)$, $v = (c_1+id_1, \ldots, c_n+id_n)$ and a scalar $\alpha$ let's prove, that $\alpha \cdot (u + v) = \alpha \cdot u + \alpha \cdot v$.

$\alpha \cdot (u + v) = \alpha \cdot \Big( (a_1+ib_1, \ldots, a_n+ib_n) + (c_1+id_1, \ldots, c_n+id_n) \Big) = \newline \alpha \cdot \Big( (a_1+c_1)+i(b_1+d_1), \ldots, (a_n+c_n)+i(b_n+d_n)\Big) = \newline \Big( \alpha \cdot (a_1+c_1)+i\alpha \cdot (b_1+d_1), \ldots, \alpha \cdot (a_n+c_n)+i\alpha \cdot (b_n+d_n)\Big) = \newline \Big( (\alpha \cdot a_1+\alpha \cdot c_1)+i(\alpha \cdot b_1+\alpha \cdot d_1), \ldots, \alpha \cdot a_n+\alpha \cdot c_n)+i(\alpha \cdot b_n+\alpha \cdot d_n) \Big) = \newline \Big( (\alpha \cdot a_1)+i(\alpha \cdot b_1), \ldots, \alpha \cdot a_n+i(\alpha \cdot b_n) \Big) + \newline \Big( (\alpha \cdot c_1)+i(\alpha \cdot d_1), \ldots, \alpha \cdot b_n+i(\alpha \cdot d_n) \Big) = \newline \alpha \cdot (a_1+ib_1, \ldots, a_n+ib_n) + \alpha \cdot (c_1+id_1, \ldots, c_n+id_n) = \alpha \cdot u + \alpha \cdot v.$

$\square$

8) DISTRIBUTIVITY ( OF SCALAR MULTIPLICATION WITH RESPECT TO FIELD ADDITION )

For any element $v = (c_1+id_1, \ldots, c_n+id_n) \in V$, and 2 scalars $\alpha$ and $\beta$ let's prove that: $(\alpha + \beta) \cdot v = \alpha \cdot v + \beta \cdot v.$

$(\alpha + \beta) \cdot v = (\alpha + \beta) \cdot (c_1+id_1, \ldots, c_n+id_n) = \newline \Big( (\alpha + \beta) \cdot (c_1+id_1), \ldots, (\alpha + \beta) \cdot (c_n+id_n) \Big) = \newline \Big( \alpha \cdot (c_1+id_1) + \beta \cdot (c_1+id_1), \ldots, \alpha \cdot (c_n+id_n) + \beta \cdot (c_n+id_n) \Big) = \newline \Big( \alpha \cdot (c_1+id_1), \ldots, \alpha \cdot (c_n+id_n) \Big) + \Big( \beta \cdot (c_1+id_1), \ldots, \beta \cdot (c_n+id_n) \Big) = \newline \alpha \cdot (c_1+id_1, \ldots, c_n+id_n) + \beta \cdot (c_1+id_1, \ldots, c_n+id_n) = \alpha \cdot v + \beta \cdot v.$

$\square$

$\blacksquare$