Find the dimension and a base of the following vector subspace:

Question

I need to find the dimension and any base of the vector subspace $A$ over a field $\Bbb{R}$: $$A=\lbrace(z_{1}, -2z{_1}, z_{1}, z_{4})\epsilon\Bbb{C}^{4}\rbrace.$$ If $A$ was a vector subspace over a field $\Bbb{C}$ I would just find the base like that:$$z=(z_{1}, -2z{_1}, z_{1}, z_{4})=z_{1}(1,-2,1,0)+z_{4}(0,0,0,1)$$ and the base would be: $$B=\lbrace(1,-2,1,0),(0,0,0,1)\rbrace,$$ but now, the scalars $z_{1}$,$z_{4}$ are not from the field $\Bbb{R}$! And I don't know how to find the base that generates the whole subspace with scalars from $\Bbb{R}$.

Answer 1

I would just find the base like that:$$z=(z_{1}, -2z{_1}, z_{1}, z_{4})=z_{1}(1,-2,1,0)+z_{4}(0,0,0,1)$$

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Okay, but how do I find it?

With $z_1 = a+bi$ and $z_4=c+di$ and $a,b,c,d \in \mathbb{R}$: $$\begin{align} z & = z_{1}(1,-2,1,0)+z_{4}(0,0,0,1) \\ & = (a+bi)(1,-2,1,0)+(c+di)(0,0,0,1) \\ & = \color{red}{a}\color{blue}{(1,-2,1,0)}+\color{red}{b}\color{blue}{i(1,-2,1,0)}+\color{red}{c}\color{blue}{(0,0,0,1)}+\color{red}{d}\color{blue}{i(0,0,0,1)} \end{align}$$ Now notice that $z$ is written as a linear combination of four (blue) vectors with real (red) coefficients. This means these four vectors span the subspace $A$ (over $\mathbb{R}$) and they form a basis if they are linearly independent (which they are, you can check).

Answer 2

$ \mathbb C^4$ as a vectorspace over $ \mathbb R$ has the dimension $8$

Thus a base you are looking for is given by

$$B=\lbrace(1,-2,1,0),(0,0,0,1), i(1,-2,1,0), i(0,0,0,1)\rbrace.$$