What will be the standard basis of $\mathbb{C}^3$ or in general how can I find the standard basis for $\mathbb{C}^n$ ?
Note: $\mathbb{C}$ is complex vector space
What will be the standard basis of $\mathbb{C}^3$ or in general how can I find the standard basis for $\mathbb{C}^n$ ?
Note: $\mathbb{C}$ is complex vector space
On
Let's start by recalling the general definition of a vector space. I report below the definition given by Kaplan [1]:
A vector space $V$ is a set of elements such that addition is defined in $V$: each two elements $\pmb{u}$, $\pmb{v}$ in $V$ have a sum $\pmb{u} + \pmb{v}$ in $V$. Element of $V$ can be multiplied by scalars (real numbers): if $c$ is a scalar and $\pmb{u}$ is in $V$, then $c\pmb{u}$ is in $V$; [...]
In order to answer your question and explain to @A_for_ Abacus the very good answer by @Brian Fitzpatrick, we need to focus our attention to the real numbers in the parenthesis above.
Then, Kaplan [1] explains the concept of a complex vector space and says:
For some applications we must change out definition of a vector space to allow the scalars to be complex numbers. One then refers to a complex vector space. To a remarkable extent, the theory or real vector spaces, as defined above, is the same as the theory of complex vector spaces.
At this point it should be clear, as @Brian Fitzpatrick has already pointed out, if $\mathbb{C}^3$ is considered as $\mathbb{C}$ vector space (so complex numbers as scalars) thus one possible basis could be: $$ \pmb{e}_1 = (1, 0, 0)\qquad \pmb{e}_2 = (0, 1, 0)\qquad \pmb{e}_3 = (0, 0, 1) $$
but if you are considering $\mathbb{C}^3$ as an $\mathbb{R}$ vector space then you get the six bases listed by @Brian Fitzpatrick.
[1] Kaplan, W., & Lewis, D. J. (1970). Calculus and Linear Algebra: Vector spaces, many-variable calculus, and differential equations (Vol. 2). Wiley.
If we are considering $\Bbb C^3$ as a $\Bbb C$-vector space, then a basis is \begin{align*} \vec e_1 &= (1,0,0) & \vec e_2 &= (0,1,0) & \vec e_3=(0,0,1) \end{align*} However, if we are considering $\Bbb C^3$ as a $\Bbb R$-vector space, then a basis is \begin{align*} \vec e_1 &= (1,0,0) & \vec e_2 &= (0,1,0) & \vec e_3=(0,0,1) \\ \vec e_4 &= (i,0,0) & \vec e_5 &= (0,i,0) & \vec e_6=(0,0,i) \end{align*} Of course this generalizes in the obvious way and \begin{align*} \dim_{\Bbb C}\Bbb C^n &= n & \dim_{\Bbb R}\Bbb C^n &= 2n \end{align*}