What is a basis for the vector space $ \Bbb{C}^{n} $ (a complex vector space)? I know that a basis for $ \Bbb{C} $ is $ \{ 1,i \} $. This set is linearly independent in $ \Bbb{C} $ and spans $ \Bbb{C} $. I think that the dimension of $ \Bbb{C}^{n} $ may be $ 2 n $, but I’m just failing to understand what kind of vectors should be in a basis.
Also, please correct me if the dimension of $ \Bbb{C}^{n} $ is not $ 2 n $.
 A: $\mathbb C$ as a vector space over $\mathbb R$ has a two-dimensional basis, e.g., $\{ 1, i \}$. However $\mathbb C$ as a vector space over $\mathbb C$ has a one-dimensional basis, such as  $\{ 1 \}$.
So the question is are you considering $\mathbb C^n$ 


*

*as a vector space over the reals, in which case $\mathbb C^n$  is a $2n$ dimensional vector space with a $2n$ member basis,  such as $\{ (1,0,...,0), (i,0,...,0), (0, 1, 0,..., 0), (0, i, 0,..., 0), ..., (0, 0, ..., 1), (0, 0, ..., i) \}$

*as a vector space over the complex numbers, in which case $\mathbb C^n$  is $n$-dimensional with a basis such as $\{ (1,0,...,0), (0, 1, 0,..., 0), ..., (0, 0, ..., 1) \}$.
There are also other more pathological possibilities, such as $\mathbb C^n$ as a vector space over the rational numbers.
A: the basis is $e_1,...,e_n,f_1,...,f_n$ such that $e_i$ is the basis for $\mathbb{R^n}$ and $f_j=i*e_j$
if you consider it over field of reals.otherwise it has n dimensions.
A: First, you have to be clear what is the field over which you want to describe it as vector space. For example $\mathbb C$ can be seen as a vector space over $\mathbb C$ (in which case the dimension is $1$ and any non-zero complex number can serve as basis, with $1$ being the canonical choice), as vector space over $\mathbb R$ (in which case the dimension is $2$ and any two non-zero complex numbers $z_1$ and $z_2$ with $\frac{z_1}{z_2}\notin\mathbb R$ give a basis, with the canonical choice being $\{1,\mathrm i\}$), or even as vector space over $\mathbb Q$ (in which case the vector space is infinite, and there's no canonical choice of basis, I don't think you can even explicitly specify one; any two incommensurable numbers are linearly independent in that case).
Given that you gave "the" basis of $\mathbb C$ as $\{1,\mathrm i\}$ I assume that you are interested in $\mathbb C^n$ as a real vector space (normally, if one says $\mathbb C^n$ without further qualification, the tacit assumption is a complex vector space).
The canonical choice for $\mathbb C^n$ as a real vector space is to use the vectors
$$\left\{\pmatrix{1\\0\\\vdots\\0},\pmatrix{\mathrm i\\0\\\vdots\\0},\pmatrix{0\\1\\\vdots\\0},\pmatrix{0\\\mathrm i\\\vdots\\0},\dots,\pmatrix{0\\0\\\vdots\\1},\pmatrix{0\\0\\\vdots\\\mathrm i}\right\}$$
(and you're right that the dimension is $2n$).
However that's not the only possible choice. The most general possibility is obtained by choosing an arbitrary basis for $\mathbb R^{2n}$ and then combining pairs of consecutive entries to a single complex number by using them as real and imaginary part. For example, for the space $\mathbb C^2$, you'd select a basis in $\mathbb R^4$, say
$$\left\{\pmatrix{a_1\\a_2\\a_3\\a_4},\pmatrix{b_1\\b_2\\b_3\\b_4},\pmatrix{c_1\\c_2\\c_3\\c_4},\pmatrix{d_1\\d_2\\d_3\\d_4}\right\}$$
and then combine them to the basis of $\mathbb C^2$ as real vector space as
$$\left\{\pmatrix{a_1+\mathrm i a_2\\a_3+\mathrm i a_4},\pmatrix{b_1+\mathrm i b_2\\b_3+\mathrm i b_4},\pmatrix{c_1+\mathrm i c_2\\c_3+\mathrm i c_4},\pmatrix{d_1+\mathrm i d_2\\d_3+\mathrm i d_4}\right\}$$
However in most cases one would not go that general, but use the following construction that still keeps some of the structure of the complex vector space visible, allowing to easily switch between $\mathbb C^n$ as real and as complex vector space:
Choose a basis $\{e_1,\ldots,e_n\}$ of $\mathbb R^n$ (note the difference in dimension to the previous case), and then use as your basis for $\mathbb C^n$ the basis $\{e_1,\mathrm i e_1, \ldots, e_n,\mathrm i e_n\}$.
Note that if you choose the canonical basis for $\mathbb R^n$, you'll get from this construction the canonical basis for $\mathbb C^n$ that I wrote above.
