# Is S = real space(R^3)

• August 25th 2010, 12:42 PM
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?
• August 25th 2010, 01:53 PM
Haven
Can you explain what you mean by real space( $\mathbb{R}^3$). I can't find that terminology in any of my algebra texts.

Are you asking if S forms a basis for $\mathbb{R}^3$?
• August 25th 2010, 08:04 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 $(2a-b,a,b,-a)$ is 4-dimensional and thus cannot belong to $\mathbb{R}^3$

Tonio
• August 26th 2010, 03:28 AM
Ackbeet
Quote:

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