ThmDex – An index of mathematical definitions, results, and conjectures.
Complex matrix determinant is homogeneous with respect to multiplying a row or a column by a constant
Formulation 0
Let $A \in \mathbb{C}^{N \times N}$ be a D6159: Complex square matrix such that
(i) $N \in \{ 1, 2, 3, \ldots \}$ is a D5094: Positive integer
(ii) $a_1, \ldots, a_N \in \mathbb{C}^{N \times 1}$ are each a D5689: Complex column matrix
(iii) $b_1, \ldots, b_N \in \mathbb{C}^{1 \times N}$ are each a D5688: Complex row matrix
(iv) \begin{equation} A = \begin{bmatrix} a_1 & \cdots & a_k & \cdots & a_N \end{bmatrix} \end{equation}
(v) \begin{equation} A = \begin{bmatrix} b_1 \\ \vdots \\ b_k \\ \vdots \\ b_N \end{bmatrix} \end{equation}
(vi) $\lambda \in \mathbb{C}$ is a D1207: Complex number
(vii) \begin{equation} X = \begin{bmatrix} a_1 & \cdots & \lambda a_k & \cdots & a_N \end{bmatrix} \end{equation}
(viii) \begin{equation} Y = \begin{bmatrix} b_1 \\ \vdots \\ \lambda b_k \\ \vdots \\ b_N \end{bmatrix} \end{equation}
Then
(1) \begin{equation} \text{Det} X = \lambda \text{Det} A \end{equation}
(2) \begin{equation} \text{Det} Y = \lambda \text{Det} A \end{equation}
Proofs
Proof 0
Let $A \in \mathbb{C}^{N \times N}$ be a D6159: Complex square matrix such that
(i) $N \in \{ 1, 2, 3, \ldots \}$ is a D5094: Positive integer
(ii) $a_1, \ldots, a_N \in \mathbb{C}^{N \times 1}$ are each a D5689: Complex column matrix
(iii) $b_1, \ldots, b_N \in \mathbb{C}^{1 \times N}$ are each a D5688: Complex row matrix
(iv) \begin{equation} A = \begin{bmatrix} a_1 & \cdots & a_k & \cdots & a_N \end{bmatrix} \end{equation}
(v) \begin{equation} A = \begin{bmatrix} b_1 \\ \vdots \\ b_k \\ \vdots \\ b_N \end{bmatrix} \end{equation}
(vi) $\lambda \in \mathbb{C}$ is a D1207: Complex number
(vii) \begin{equation} X = \begin{bmatrix} a_1 & \cdots & \lambda a_k & \cdots & a_N \end{bmatrix} \end{equation}
(viii) \begin{equation} Y = \begin{bmatrix} b_1 \\ \vdots \\ \lambda b_k \\ \vdots \\ b_N \end{bmatrix} \end{equation}
Using R5517: Cofactor partition for a complex square matrix, we have \begin{equation} \text{Det} Y = \sum_{n = 1}^N Y_{k, n} C_{k, n} = \sum_{n = 1}^N \lambda A_{k, n} C_{k, n} = \lambda \sum_{n = 1}^N A_{k, n} C_{k, n} = \lambda \text{Det} A \end{equation} The case for $X$ (where a column is scaled) is proven anologously using the same result. $\square$