Is S = real space(R^3)

**experiment00** · August 25th 2010, 11:42 AM

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

how do prove and show this one?

**Haven** · August 25th 2010, 12:53 PM

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 $\mathbb{R}^3$ ?

**tonio** · August 25th 2010, 07:04 PM

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

**Ackbeet** · August 26th 2010, 02:28 AM

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.

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