# Is S = real space(R^3)

• Aug 25th 2010, 11:42 AM
experiment00
Is S = real space(R^3)
where S = {[ 2a-b, a , b , -a ] | a,b are real numbers}

how do prove and show this one?
• Aug 25th 2010, 12:53 PM
Haven
Can you explain what you mean by real space(\$\displaystyle \mathbb{R}^3\$). I can't find that terminology in any of my algebra texts.

Are you asking if S forms a basis for \$\displaystyle \mathbb{R}^3\$?
• Aug 25th 2010, 07:04 PM
tonio
Quote:

Originally Posted by experiment00
where S = {[ 2a-b, a , b , -a ] | a,b are real numbers}

how do prove and show this one?

The vector \$\displaystyle (2a-b,a,b,-a)\$ is 4-dimensional and thus cannot belong to \$\displaystyle \mathbb{R}^3\$

Tonio
• Aug 26th 2010, 02:28 AM
Ackbeet
Quote:

The vector \$\displaystyle (2a-b,a,b,-a)\$ is 4-dimensional and thus cannot belong to \$\displaystyle \mathbb{R}^{3}.\$
In addition, it is a vector with two degrees of freedom (it has two parameters, a and b). And so, considered as a subset of \$\displaystyle \mathbb{R}^{4},\$ it is two-dimensional at most.