Answer to Question #140944 in Linear Algebra for Alok

Question

in the following a set V, a field F, which is either R or C, and operations of addition + and scalar multiplication . , are given for alpha element of F and x element of V, we write their multiplication alpha × x as alpha • x, cheak whether V is a vector space over F, with these operations .

Answer 1

As I understand it, we want to check 10 axioms of a vector space.

1) AC Additive Closure

$$\text{if}\quad \vec{u},\vec{v}\in V, \quad\text{then}\quad \left(\vec{u}+\vec{v}\right)\in V$$

2) SC Scalar Closure

$$\text{if}\quad\alpha\in\mathbb{C}\quad\text{and}\quad \vec{u}\in V,\quad\text{then}\quad(\alpha\cdot\vec{u})\in V$$

3) C Commutativity

$$\text{if}\quad \vec{u},\vec{v}\in V, \quad\text{then}\quad \vec{u}+\vec{v}=\vec{v}+\vec{u}$$

4) AA Additive Associativity

$$\text{if}\quad \vec{u},\vec{v},\vec{w}\in V, \quad\text{then}\quad \left(\vec{u}+\vec{v}\right)+\vec{w}=\vec{v}+\left(\vec{u}+\vec{w}\right)$$

5) Z Zero Vector

$$\text{There is a vector},\,\,\, \vec{0},\,\,\,\text{called the zero vector, such that}\\[0.3cm]\,\,\,\vec{u}+\vec{0}=\vec{u}\,\,\,\text{for all}\,\,\,\vec{u}\in V$$

6) AI Additive Inverses

$$\text{If}\,\,\,\vec{u}\in V,\,\,\,\text{then there exists a vector}\,\,\,(−\vec{u})\in V\,\,\,\text{so that}\\[0.3cm] \vec{u}+(−\vec{u})=\vec{0}.$$

7) SMA Scalar Multiplication Associativity

$$\text{If} \,\,\,\alpha, \beta\in\mathbb{C}\,\,\,\text{and}\,\,\,\vec{u}\in V, \,\,\,\text{then}\\[0.3cm] \alpha\cdot(\beta\cdot\vec{u})=(\alpha\beta)\cdot\vec{u}$$

8) DVA Distributivity across Vector Addition

$$\text{If}\,\,\,\alpha\in\mathbb{C}\,\,\,\text{and}\,\,\,\vec{u},\vec{v}\in V,\,\,\,\text{then}\\[0.3cm] \alpha\cdot(\vec{u}+\vec{v})=\alpha\cdot\vec{u}+\alpha\cdot\vec{v}$$

9) DSA Distributivity across Scalar Addition

$$\text{If}\,\,\,\alpha,\beta\in\mathbb{C}\,\,\,\text{and}\,\,\,\vec{u}\in V,\text{then}\\[0.3cm] (\alpha+\beta)\cdot\vec{u}=\alpha\cdot\vec{u}+\beta\cdot\vec{u}$$

10) O One

$$\text{If}\,\,\,\vec{u}\in V,\,\,\,\text{then}\,\,\, 1\cdot\vec{u}=\vec{u}.$$