HELLO,
Is a subspace of $\displaystyle {C^3}$?
C=complex numbers.
Thanks
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.
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.