# Thread: Is R^3 a subspace of C^3

1. ## Is R^3 a subspace of C^3

HELLO,

Is a subspace of $\displaystyle {C^3}$?

C=complex numbers.

Thanks

2. Is it closed under addition? yes
Is it closed under scalar multiplication from the field C? no
$\displaystyle <1,1,1> \in \mathbb{R}^3$ but $\displaystyle i<1,1,1>=<i,i,i> \notin \mathbb{R}^3$
Not a subfield.

3. How do i prove closed under addition condition.
you took i from complex vector space C. Can you do

$\displaystyle ((1+i),1,1) \notin \mathbb{R}^3$

4. well vector addition is just done componentwise, so if you add two real valued vectors, their sum is clearly still real valued since the real numbers are closed under addition. I mean (a,b,c)+(d,e,f)=(a+d,b+e,c+f) where everything is necessarily real numbers, so its closed under addition. But to be a subspace it also needs to be closed under scalar multiplication, and it is not, this is enough to prove it is not a subspace.

5. I think it depends on the field.

For example: Every field is a one dimentional vector space over itself.

This implies that $\displaystyle \mathbb{C}$ has A basis $\displaystyle \mathcal{B}=\{ 1\}$ when the scalers are take as complex numbers.

On the other hand $\displaystyle \mathbb{C}$ has A basis $\displaystyle \mathcal{B'}=\{ 1,i\}$ when the scalers are taken as real numbers. It is a two dimentional vectors space over $\displaystyle \mathbb{R}$

Now we could view $\displaystyle \mathbb{R}$ as a closed subspace of the 2nd, but not the first.