$\mathbb{C}^{n}$ vector space, $\mathbb{R}$ field, $\det(A) = 0 + bi$, $b \neq 0$, does this imply columns of the matrix are linearly independent?

Question

Given a subset of $n$ vectors of the vector space $\mathbb{C}^n$ and form an n x n matrix, call it $A$. If $\det(A) = 0 + bi$, where $b \in \mathbb{R}\setminus\left\{0\right\}$. Would this imply that the $n$ vectors are linearly independent when we consider $\mathbb{C}^{n}$ as a vector space over the field $\mathbb{R}$?

Answer 1

Well, for $Ae_i$ (the columns of $A$) to be linearly independent over $\mathbb{R}$, we must have that if $\sum_{i=1}^n x_i (Ae_i) = \vec{0}$ with $x_i \in \mathbb{R}$ then $x_1 = \dots = x_n = 0$. But

$$ \sum_{i=1}^n x_i A(e_i) = \sum_{i=1}^n A(x_i e_i) = A \begin{pmatrix} x_1 \\ \vdots \\ x_n \end{pmatrix} = A\vec{x}. $$

Since $\det(A) \neq 0$, the matrix $A$ has trivial kernel (over $\mathbb{C}^n$) and in particular, if $\vec{x} \in \mathbb{R}^n$ such that $A\vec{x} = \vec{0}$ then $\vec{x} = \vec{0}$.