Linear independence of complex basis of vectors. I understand that for a $\mathbb{C}^n$ as a real vector space, we choose $$\left\{\pmatrix{1\\0\\\vdots\\0},\pmatrix{\mathrm i\\0\\\vdots\\0},\pmatrix{0\\1\\\vdots\\0},\pmatrix{0\\\mathrm i\\\vdots\\0},\dots,\pmatrix{0\\0\\\vdots\\1},\pmatrix{0\\0\\\vdots\\\mathrm i}\right\}$$ as a standard basis. Now, how would I show linear independence of the basis? My first thought was to show that the determinant of the basis matrix is non-zero, but when looking at this basis I realise that there is no determinant because it is not a square matrix. 
 A: If you see $\mathbb C^n$ as a real vector space, you have the following isomorphism:
\begin{gather}
\mathbb C^n \rightarrow \mathbb R^n\times \mathbb R^n \rightarrow \mathbb R^{2n}, \qquad \pmatrix{a_1+ i b_1\\ a_2+ib_2\\ \vdots \\ a_n+ib_n}\longmapsto \left(\pmatrix{a_1\\ a_2\\ \vdots \\ a_n},\pmatrix{b_1\\ b_2\\ \vdots \\ b_n}\right)\longmapsto \pmatrix{a_1 \\ b_1\\ a_2 \\ b_2\\ \vdots \\ a_n \\ b_n}
\end{gather}
Now we can compute the determinant in $\mathbb R^{2n}$: the matrix is the identity matrix so they are linearly independent.
A: Let the $e_j$ be the standard basis elements for $\Bbb R^n$, so those of $\Bbb C^n$ are $e_j,\,ie_j$. A general linear combination thereof is $\sum_j(a_j+ib_j)e_j$ with $a_j,\,b_j\in\Bbb R$. If this vanishes, $a_k+ib_k=0\cdot e_k=0$, so $a_k=b_k=0$.
A: See a little case for this, for example, to show the independence of
$$S = \left\{
\begin{pmatrix} 1 \\ 0 \end{pmatrix}, \begin{pmatrix} i \\ 0 \end{pmatrix},
\begin{pmatrix} 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 0 \\ i \end{pmatrix}
\right\}$$
suppose that $\begin{pmatrix} 0 \\ 0 \end{pmatrix}$ can be written as
$$
\begin{pmatrix} 0 \\ 0 \end{pmatrix} = 
a_1\begin{pmatrix} 1 \\ 0 \end{pmatrix} + b_1\begin{pmatrix} i \\ 0 \end{pmatrix} +
a_2\begin{pmatrix} 0 \\ 1 \end{pmatrix} + b_2\begin{pmatrix} 0 \\ i \end{pmatrix}
$$
for some choice of real numbers $a_1,b_1,a_2$ and $b_2$. Observe that the right hand side of this is
$$\begin{pmatrix} a_1+ib_1 \\ a_2 + ib_2 \end{pmatrix}$$
hence $a_1+ib_1 = 0$ and $a_2 + ib_2 = 0$. Of course, the latter implies that $a_1=b_1=a_2=b_2=0$, which means that the only way of writting $\begin{pmatrix} 0 \\ 0 \end{pmatrix}$ as a linear combination of the vectors in $S$ is the trivial combination. Thus, $S$ is linearly independent.
