Suppose $X$ is a vector space over $\mathbb C$ and has as basis $\{e_1,e_2,\ldots,e_n\}$. Now regard $X$ as a vector space over $\mathbb R$.

What will be the basis?

My thoughts:

I considered $\mathbb C$ over $\mathbb C$ and $\mathbb C$ over $\mathbb R$.In the first case we have $(1,0)$ as basis and in the latter case we have $\{(1,0),(0,1)\}$ as basis i.e. $\{(1,0),(0,1)(1,0)=(0,1)\}$ as basis.

So may be the answer is $\{e_1,e_2,\ldots,e_n,ie_1,\ldots,ie_n\}$. How to justify the result if its true?

  • 1
    $\begingroup$ Yes, your answer is correct. As usual t's enough to show that (1) the basis is linearly independent (over $\Bbb R$), and (2) that any element of $\Bbb C^n$ can be written as an ($\Bbb R$-)linear combination of basis elements. $\endgroup$ – Travis Aug 3 '15 at 18:12
  • $\begingroup$ What about taking Re$e_k$, Im$e_k$? $\endgroup$ – A.Γ. Aug 3 '15 at 18:14
  • $\begingroup$ @A.G. That doesn't work. Consider the standard basis. Half of your basis elements are $0$. $\endgroup$ – user24142 Aug 3 '15 at 18:31
  • $\begingroup$ @user24142 That's right, I see now, thanks! $\endgroup$ – A.Γ. Aug 3 '15 at 18:41
  • $\begingroup$ And what does "$\operatorname{Re} e_k$" means when $e_k$ is a vector in a complex vector space? $\endgroup$ – Najib Idrissi Aug 3 '15 at 18:51

In the case of $\mathbb{C}$ over $\mathbb{C}$, the basis would be $\{1\}$ because every element of $\mathbb{C}$ can be written as a $\mathbb{C}$-multiple of $1$.

$$\mathbb{C}=\{z\times 1 : z \in \mathbb{C}\}$$

In the case of $\mathbb{C}$ over $\mathbb{R}$, the basis would be $\{1,\mathrm{i}\}$ because every element of $\mathbb{C}$ can be written as an $\mathbb{R}$-multiple of $1$ and $\mathrm{i}$.

$$\mathbb{C}=\{x\times 1 + y \times \mathrm{i} : x,y \in \mathbb{R}\}$$

If $\{{\bf v}_1,\ldots,{\bf v}_n\}$ is a basis for $V$ over $\mathbb{C}$ then $$V = \{a_1{\bf v}_1+\cdots+a_n{\bf v}_n: a_k \in \mathbb{C} \}$$

We can write each of the $a_k$ as $b_k+\mathrm{i}c_k$, where $b_k,c_k \in \mathbb{R}$. Hence \begin{eqnarray*} a_1{\bf v}_1+\cdots+a_n{\bf v}_n &=& (b_1+\mathrm{i}c_1){\bf v}_1+\cdots+(b_n+\mathrm{i}\mathrm{c}_n){\bf v}_n \\ &=& b_1{\bf v}_1+\cdots+b_n{\bf v}_n+c_1(\mathrm{i}{\bf v}_1)+\cdots+c_n(\mathrm{i}{\bf v}_n) \end{eqnarray*}

We can take $\{{\bf v}_1,\ldots,{\bf v}_n,\mathrm{i}{\bf v}_1,\ldots,\mathrm{i}{\bf v}_n\}$ as a basis for $V$.

The final step is to show that

$$V = \mathbb{R}\langle {\bf v}_1,\ldots,{\bf v}_n\rangle \oplus \mathbb{R}\langle \mathrm{i}{\bf v}_1,\ldots,\mathrm{i}{\bf v}_n\rangle$$

This is obvious since $\mathbb{i} \notin \mathbb{R}$.


Suppose $x\in X$. Then $x = c_1 e_1 + \cdots+c_n e_n$ for some complex numbers $c_1,\ldots,c_n$.

For $k=1,\ldots,n$ write $c_k = a_k + i b_k$ where $a_k$ and $b_k$ are real.

Then \begin{align} x & = c_1 e_1 + \cdots+c_n e_n \\[8pt] & = (a_1+ib_1)e_1 + \cdots + (a_n+ib_n)e_n \\[8pt] & = a_1 e_1 + \cdots + a_n e_n + b_1(ie_1) + \cdots + b_n (ie_n). \end{align} So $x$ is a linear combination of $e_x,\ldots,e_n,ie_1,\ldots,ie_n$ with coefficients that are real.

Linear independence can be proved by considering almost the same sequence of equalities: \begin{align} 0 & = a_1 e_1 + \cdots + a_n e_n + b_1(ie_1) + \cdots + b_n (ie_n) \tag 1 \\[8pt] & = (a_1+ib_1)e_1 + \cdots + (a_n+ib_n)e_n \\[8pt] & = c_1 e_1 + \cdots+c_n e_n \end{align} and that can be true only if $c_1=\cdots=c_n=0$, by linear independence of $e_1,\ldots,e_n$ over $\mathbb C$. Hence $(1)$ can be true only if $a_1=\cdots=a_n=b_1=\cdots=b_n=0$.

  • $\begingroup$ Could the person who down-voted this explain why? ${}\qquad{}$ $\endgroup$ – Michael Hardy Aug 3 '15 at 22:49

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