I am a bit rusty on Vector Spaces; please verify my answers to the following questions:
1.
Let $V=\mathbb R^4$ and $v_1=(2,0,3,1), v_2=(2,3,0,1), v_3=(2,1,2,1)$.
Does $v=(x_1,x_2,x_3,x_4)\in span(v_1,v_2,v_3)$ if $x_1=2x_4,x_3=-x_2+3x_4$?
Solution Outline: $v=(2x_4,x_2,-x_2+3x_4,x_4)=x_4(2,0,3,1)+x_2(0,1,-1,0)$ So if I show that both $(2,0,3,1)$ and $(0,1,-1,0)$ belong to $span(v_1,v_2,v_3)$, then I am done — is that right? Notice that $(2,0,3,1)=v_1$
2.
Basis for $W=\{a+(a+b)x+(2a+b)x^2+bx^3| \; a,b\in \mathbb R\}$
if $\alpha\in W\Rightarrow \text{For some a,b in R} \; \alpha = a+(a+b)x+(2a+b)+bx^3 = a(1+x+2x^2) +b(x+x^2+x^3)$
Hence $W=span(1+x+2x^2,x+x^2+x^3)$, which are also linearly independent, so this is the required basis.
3.
Is $Span(\frac{1+i\sqrt3}{2},\frac{1-i\sqrt3}{2}) = Span(1,\sqrt3 )$ over $\mathbb R $ and $\mathbb C$?
According to me, Over $\mathbb R$, we basically want to check if $Span((\frac12,\frac{\sqrt3}{2}),(\frac12,\frac{-\sqrt{3}}{2})) = Span((1,0),(\sqrt3,0))$, which is true.
Over $\mathbb C$, I don't think $\frac{1+i\sqrt3}{2}$ belongs to $Span(1,\sqrt3 )$, so they can't be equal. Does this makes sense?