Let $A \in \mathbb{C}^{N \times N}$ be a
D6159: Complex square matrix such that
| (i) |
$a_1, \ldots, a_N \in \mathbb{C}^{N \times 1}$ are each a D5689: Complex column matrix
|
| (ii) |
$b_1, \ldots, b_N \in \mathbb{C}^{1 \times N}$ are each a D5688: Complex row matrix
|
| (iii) |
\begin{equation}
A
=
\begin{bmatrix}
a_1 & a_2 & \cdots & a_N
\end{bmatrix}
\end{equation}
|
| (iv) |
\begin{equation}
A
=
\begin{bmatrix}
b_1 \\
b_2 \\
\vdots \\
b_N
\end{bmatrix}
\end{equation}
|
If $N = 1$, the case is clear, so we can assume that $N \geq 2$. Suppose that $a_k = \boldsymbol{0}$ for some $k \in \{ 1, \ldots, N \}$. Using
R5517: Cofactor partition for a complex square matrix, we have
\begin{equation}
\text{Det} A
= \sum_{n = 1}^N A_{k, n} C_{k, n}
= \sum_{n = 1}^N 0 \cdot C_{k, n}
= 0
\end{equation}
If instead $b_k = \boldsymbol{0}$, then the same result implies
\begin{equation}
\text{Det} A
= \sum_{n = 1}^N A_{n, k} C_{n, k}
= \sum_{n = 1}^N 0 \cdot C_{n, k}
= 0
\end{equation}
Since $k \in \{ 1, \ldots, N \}$ was arbitrary, we are done. $\square$
