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. 

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 $a_k=b_k=0$.
