 # Describe the zero vector (the additive identity) of the vector space. – Given Vector Space:

$\mathbb{R}^4$ Real number

Additive Identity is defined as the value that if added or subtracted from a second value, does not change it. For example, if we add $0$ to any real numbers, it does not change the value of the given real numbers. We can call Zero $0$ the Additive Identity of the Real Numbers.

If we consider $R$ as a real number and $I$ as an Additive Identity, then as per Additive Identity Law:

$R+I=I+R=R$ Subtraction real number

A Vector Space is defined as a Set comprising of one or more vector elements and it is represented by $\mathbb{R}^n$ where $n$ represents the number of elements in the given vector space. Given that:

Vector Space $=\mathbb{R}^4$

This shows that $\mathbb{R}^4$ has $4$ vector elements.

Let us represent $\mathbb{R}^4$ as follows:

$\mathbb{R}^4 =\ (R_1,\ R_2,\ R_3,\ R_4)$

Let us suppose that:

Additive Identity $=\mathbb{I}^4$

Let us represent $= \mathbb{I}^4$ as follows:

$\mathbb{I}^4 = (I_1,\ I_2,\ I_3,\ I_4)$

$\mathbb{R}^4\ +\mathbb{I}^4\ =\mathbb{I}^4\ +\mathbb{R}^4\ =\ \mathbb{R}^4$

Substituting the values:

$(R_1,\ R_2,\ R_3,\ R_4)\ +\ (I_1,\ I_2,\ I_3,\ I_4)\ =\ (R_1,\ R_2,\ R_3,\ R_4)$

$(R_1\ +\ I_1,\ R_2\ +{\ I}_2,\ R_3\ +{\ I}_3,\ R_4{\ +\ I}_4)\ =\ (R_1,\ R_2,\ R_3,\ R_4)$

Comparing element by element:

First Element:

$R_1\ +{\ I}_1\ =\ R_1$

$I_1\ =\ R_1\ -{\ R}_1$

$I_1\ =\ 0$

Second Element:

$R_2\ +\ I_2\ ={\ R}_2$

$I_2\ ={\ R}_2\ -{\ R}_2$

$I_2\ =\ 0$

Third Element:

$R_3\ +\ I_3\ =\ R_3$

$I_3\ =\ R_3\ -\ R_3$

$I_3\ =\ 0$

Fourth Element:

$R_4\ +\ I_4\ ={\ R}_4$

$I_4\ =\ R_4\ -\ R_4$

$I_4\ =\ 0$

Hence from the above equations, it is proved that the Additive Identity is as follows:

$(I_1,\ I_2,\ I_3,\ I_4)\ =\ (0,\ 0,\ 0,\ 0)$

$\mathbb{I}^4\ =\ (0,\ 0,\ 0,\ 0)$

## Numerical Result

The Additive Identity or Zero Vector $\mathbb{I}^4$ of $\mathbb{R}^4$ is:

$\mathbb{I}^4\ =\ (0,\ 0,\ 0,\ 0)$

## Example

For the given vector space $\mathbb{R}^2$, find the zero vector or additive identity.

Solution

Given that:

Vector Space $= \mathbb{R}^2$

This shows that $\mathbb{R}^2$ has $2$ vector elements.

Let us represent $\mathbb{R}^2$ as follows:

$\mathbb{R}^2\ =\ (R_1,\ R_2)$

Let us suppose that:

Additive Identity $= \mathbb{I}^2$

Let us represent $= \mathbb{I}^2$ as follows:

$\mathbb{I}^2\ =\ (I_1,\ I_2)$

$\mathbb{R}^2\ +\ \mathbb{I}^2\ =\ \mathbb{I}^2\ +\ \mathbb{R}^2\ =\ \mathbb{R}^2$

Substituting the values:

$(R_1,\ {\ R}_2)\ +\ (I_1,\ \ I_2)\ =\ (R_1,\ R_2)$

$(R_1\ +{\ I}_1,\ \ R_2\ +\ I_2)\ =\ (R_1,\ R_2)$

Comparing element by element:

First Element:

$R_1\ +{\ I}_1\ =\ {\ R}_1$

$I_1\ ={\ R}_1\ -{\ R}_1$

$I_1\ =\ 0$

Second Element:

$R_2\ +\ I_2\ ={\ R}_2$

$I_2\ ={\ R}_2\ -{\ R}_2$

$I_2\ =\ 0$

Hence from the above equations, it is proved that the Additive Identity is as follows:

$(I_1,\ {\ I}_2)\ =\ (0,\ 0)$

$\mathbb{I}^2\ =\ (0,\ 0)$