If $V=\overline V$ where $V$ is a vector space, then there exists a real basis of $V$

Question

Considering $V$ to be a vector space over $\mathbb C$ then if $V=\overline V$ how can i assure there is a real basis?

My first consideration was that if $w \in V$ then $\bar w \in V$ so $Re(w) \in V$ but i can't construct a basis over this fact.

Given a basis i know that its conjugated is also a basis, is the sum of elements of a basis also a basis?

Answer 1

When you say $V=\overline{V}$, I'll assume that what you mean is there exists an additive involution of $V$, $v\mapsto \bar{v}$, such that $\overline{zv} = \bar{z}\bar{v}$. The real subspace is then the subset $W=\renewcommand\Re{\operatorname{Re}}\Re(V)=\{v\in V: v=\bar{v}\}$.

We want to construct a complex basis of $V$ which lies in $W$.

Observe that as real vector spaces $V=W\oplus iW$, since for any $v\in V$, $$v = \frac{v+\bar{v}}{2} + i \frac{v-\bar{v}}{2i},$$ and $W\cap iW=0$.

Then choose a real basis for $W$, $v_1,\ldots,v_n$. I claim $v_1,\ldots,v_n$ is a $\Bbb{C}$ basis for $V$.

To see this, simply observe that $\dim_{\Bbb{R}} W = \frac{1}{2}\dim_{\Bbb{R}} V = \dim_{\Bbb{C}}V$, since $V=W\oplus iW$, and $v_1,\ldots,v_n,iv_1,\ldots,iv_n$ span $V$ as an $\Bbb{R}$ vector space. Therefore $v_1,\ldots,v_n$ span $V$ as a $\Bbb{C}$ vector space, and they must form a basis, since there are the correct number of vectors.

Special case

In the particular case where we start with a real vector space $W$, and define $V=W\otimes_{\Bbb{R}} \Bbb{C}$, the involution on $V$ is induced by complex conjugation on $\Bbb{C}$, and the real subspace can naturally be identified with $W$ via the map $w\mapsto w\otimes 1$.

In the even more particular case where $W=\Bbb{R}^n$, $V=\Bbb{C}^n$, the involution on $V$ induced by complex conjugation is exactly what you'd expect, namely the conjugate of a column vector $$\begin{pmatrix} z_1 \\ z_2 \\ \vdots\\z_n\end{pmatrix}\text{ is }\begin{pmatrix} \bar{z_1} \\ \bar{z_2} \\ \vdots\\\bar{z_n}\end{pmatrix}.$$

Answer 2

I'm going to assume that $\bar{V}=\{\bar{v}\in\mathbb{C}\mid v\in V\}$.

In which case, by definition...

$$V=\mathbb{R}\otimes B$$

Where $B\subset\mathbb{C}$ is your basis.

For each $z\in B$, there are real numbers $a$ and $b$ such that $z=a+bi$. Then, again by definition $\bar{z}=a-bi$. It is trivial to show that $\mathbb{R}=\bar{\mathbb{R}}$, so...

$$V=\bar{V}\iff B=\bar{B}$$

Which implies...

$$\forall a+bi\in B.a+bi=a-bi\implies b=0$$

In other words, the imaginary part of your basis is always $0$.