Question

Question:If $n⩾1$ is an integer, show that $F^n$ is a vector space over F .

Short answer

Answer:

$F^n$ is a vector space over F.

Step-by-step solution

Definition of vector.

A space consisting of vector, together with the associative and commutative operations of addition of vectors and the associative and distributive operation of multiplication of vectors by scalars.

Showing the property of vector space. 

If $n⩾1$ is an integer.

The tuples can be written as a row or as a coloum vector.

Let $x=\left(x_1,x_2,.........x_n\right),y=\left(y_1,y_2,.......y_n\right),z=\left(z_1,z_2,.......z_n\right)^I{F}^n$and $a,b\in F$.

If $F^n$ is a vector space over, then it will satisfy all the condition

Commutativity of addition

For all$x=\left(x_1,x_2,.......x_n\right),y=\left(y_1,y_2,.......y_n\right)^I{F}^n$

$$\begin{gathered} x+y=\left(x_1,x_2.....x_n\right)+\left(y_1,y_2,.....y_n\right) \\ =\left(x_1+y_1,x_2+y_2,.......x_n+y_n\right) \\ =\left(y_1+x_1,y_2+x_2,......y_n+x_n\right) \\ =\left(y_1,y_2,.....\right)+\left(x_1,x_2,......\right)\to y+x \end{gathered}$$.

This gives $x+y=y+x$.

Identity element of addition.

Let then

$$\begin{gathered} 0+x=0+\left(x_1,x_2,....x_n\right) \\ =\left(x_1,x_2.......x_n\right) \\ =x \end{gathered}$$

Hence there exist $0\in F$a Such that 0+x =x for all $x\in F^n$.

Thus 0 is the identity element of the addition.

Inverse element of addition.

localid="1660143976964" $-x=\left(-x_1,-x_2,......-x_n\right)\in F^n$Then

$$\begin{gathered} -x+x=\left(-x_1,-x_2,......-x_n\right)+\left(x_1,x_2,....x_n\right) \\ =\left(0,0,0.........0\right) \\ =0 \end{gathered}$$

Step 3:Showing the property of vector space.

Scalar multiplication

$x=\left(x_1,x_2,.....\right)\in F^n$ and $a,b\in F$

$$\begin{gathered} a\left(bx\right)=a\left(b\left(x_1,x_2,.....x_n\right)\right) \\ =a\left(b{x}_1,b{x}_2,......b{x}_n\right) \\ =ab\left(x_1,x_2,.......x_n\right) \end{gathered}$$

This gives $a\left(bx\right)=\left(ab\right)x$

Identity element of scalar multiplication

Let $I\in F$ then,

$$\begin{gathered} I·x=I·\left(x_1,x_2,......x_n\right) \\ =\left(x_1,x_2,......x_n\right) \\ =x \end{gathered}$$

Hence there exist $I\in F$ such that $I·x=x$ for all $x\in F^n$.

Step 4:Showing the property of vector space. 

Distributive of scalar multiplication with respect to vector addition,

Let $x=\left(x_1,x_2,....\right)\in F^{n}anda,b\in F$

$$\begin{gathered} d\left(a+b\right)\left(x\right)=\left(a+b\right)\left(x_1,x_2,....x_n\right) \\ =\left(a\left(x_1,x_2,.....\right)+b\left(x_1,x_2,....\right)\right) \end{gathered}$$

This gives,

$\left(a+b\right)\left(x\right)=ax+bx$

Since $F^n$ satisfy all above properties.

Therefore, set $F^n$ is vector space over F.