Determinant of sum of orthogonal matrix with rank-$1$ matrix

Question

What is the determinant of the sum of two matrices

$$\det (G + S)$$

where $S$ is all zeros except for a single column of $1$'s?

$$S = \begin{bmatrix} 0 & ... & 0 & 1 & 0 & ... & 0 \\ 0 & ... & 0 & 1 & 0 & ... & 0 \\ \vdots & & \vdots & \vdots & \vdots & & \vdots \\ 0 & ... & 0 & 1 & 0 & ... & 0 \\ \end{bmatrix}$$

I understand this can be solved by breaking up the determinant into columns, but I am unsure of how to do this. Also, $S$ is clearly singular - is there a general rule for the determinant of the sum of a singular and non-singular matrix (i.e. $G$ orthogonal $S$ singular)? Any help greatly appreciated

Im essentially asking what happens to the determinant of a matrix when you add $1$ to each entry in a column. Specifically, I am interested in the case where $G$ is orthogonal with $\det(G) = -1$. I would also be interested in the case where we add $1$ to just a single entry.

Answer 1

You can use the following two properties of determinants to answer this: \begin{align} &1.&\mathrm{det}\left[a_1,a_2,...,a_k+v_k,...,a_n\right] &= \mathrm{det}\left[a_1,a_2,...,a_k,...,a_n\right] + \mathrm{det}\left[a_1,a_2,...,v_k,...,a_n\right] \\ &2.&\mathrm{det}\left[a_1,a_2,...,\lambda a_k,...,a_n\right] &= \lambda \mathrm{det}\left[a_1,a_2,...,a_k,...,a_n\right] \end{align} where $a_i$, $v_k$ are columns of appropriate sizes and $\lambda \in \mathbb{R}$. Now you can generalize your problem to the following question:

Assume $G$ is orthogonal with $\mathrm{det}\left(G\right) = -1$, and let $g_i$ be the columns of $G$ then how do we compute $\mathrm{det}\left[g_1,g_2,...,g_k+v_k,...,g_n\right]$, where $v_k$ is some constant vector?

First, since $G$ is orthogonal, there exists $\alpha_i$ such that $v_k = \sum^{n}_{i=1} g_i \alpha_i$. Now you can use the first property and get: \begin{align} \mathrm{det}\left[g_1,g_2,...,g_k+v_k,...,g_n\right] &= \mathrm{det}\left[g_1,g_2,...,g_k+\sum^{n}_{i=1} g_i \alpha_i,...,g_n\right] \\ &=\mathrm{det}\left[g_1,g_2,...,\sum^{n}_{i=1,i\neq k} g_i \alpha_i,...,g_n\right]+\mathrm{det}\left[g_1,g_2,...,(1+\alpha_k)g_k,...,g_n\right]\\ &= (1+\alpha_k)\mathrm{det}\left[g_1,g_2,...,g_k,...,g_n\right]\\ &=-(1+\alpha_k) \end{align} You can also express $\alpha_k$ using orthogonal projections, $\alpha_k = \left(g_k^\mathrm{T}g_k\right)^{-1}g_k^\mathrm{T}v_k$ and thus get \begin{align} \mathrm{det}\left[g_1,g_2,...,g_k+v_k,...,g_n\right] &= -1-\left(g_k^\mathrm{T}g_k\right)^{-1}g_k^\mathrm{T}v_k \end{align}

In your case $v_k$ consists only of ones as elements. Furthermore, this discussion can also be extended to general matrices $A$ of course, but since orthogonal matrices were your concern, this is shorter.

Answer 2

Suppose that we have an invertible matrix $\mathrm A \in \mathbb R^{n \times n}$. Using Sylvester's determinant identity,

$$\det ( \,\mathrm A + 1_n \mathrm e_i^\top ) = \det(\mathrm A) \cdot \det ( \mathrm I_n + \mathrm A^{-1} 1_n \mathrm e_i^\top ) = \det(\mathrm A) \cdot ( 1 + \mathrm e_i^\top \mathrm A^{-1} 1_n )$$

If $\rm A$ is orthogonal, then $\det (\mathrm A) = \pm 1$ and $\mathrm A^{-1} = \mathrm A^\top$. Hence,

$$\det ( \,\mathrm A + 1_n \mathrm e_i^\top ) = \pm ( 1 + \mathrm e_i^\top \mathrm A^\top 1_n ) = \pm ( 1 + 1_n^\top \mathrm A \, \mathrm e_i ) = \color{blue}{\pm ( 1 + 1_n^\top \mathrm a_i )}$$

where $\mathrm a_i$ is the $i$-th column of $\rm A$. Thus,

$$\det ( \,\mathrm A + \mathrm e_i \mathrm e_j^\top ) = \pm ( 1 + \mathrm e_j^\top \mathrm A^\top \mathrm e_i ) = \pm ( 1 + \mathrm e_i^\top \mathrm A \, \mathrm e_j ) = \color{blue}{\pm ( 1 + a_{ij} )}$$