Prove that the determinant of a rank-1 matrix plus a small-norm matrix is nearly zero?

Question

Definition. Given two square matrices ${{\bf{A}}_1},{{\bf{A}}_2} \in {\mathbb{C}^{n \times n}}$, where $\left\| {{{\bf{A}}_1}} \right\| \gg \left\| {{{\bf{A}}_2}} \right\|$, $\left\| {{{\bf{A}}_2}} \right\| \ll 1$ and ${\rm{rank}}({{\bf{A}}_1}) = 1,{\rm{rank}}({{\bf{A}}_2}) = n$.

Tests show that the determinant $\left| {\det ({{\bf{A}}_1} + {{\bf{A}}_2})} \right| \approx 0$, perhaps because the sum matrix is a "nearly rank-1" matrix?

Question. I'm wondering how to prove the above observation $\left| {\det ({{\bf{A}}_1} + {{\bf{A}}_2})} \right| \approx 0$? And how small $\left\| {{{\bf{A}}_2}} \right\|$ should be to make $\left| {\det ({{\bf{A}}_1} + {{\bf{A}}_2})} \right| < \varepsilon $, where $\varepsilon $ is a predefined value?

My failed attempts:

1. According to the Matrix determinant lemma, Express ${{\bf{A}}_1}$ as ${{\bf{A}}_1} = {\bf{u}}{{\bf{v}}^{\rm{T}}}$.

2. Then $\det ({{\bf{A}}_1} + {{\bf{A}}_2}) = \det ({\bf{u}}{{\bf{v}}^{\rm{T}}} + {{\bf{A}}_2}) = (1 + {{\bf{v}}^{\rm{T}}}{{\bf{A}}_2}^{ - 1}{\bf{u}})\det ({{\bf{A}}_2})$.

3. Although $\det ({{\bf{A}}_2}) \ll 1$, the inverse ${{\bf{A}}_2}^{ - 1}$ may be very large. Thus, the above formula seems unable to serve the task.

Thanks!

Answer 1

Since $A_1$ has rank 1, $\mathbb C^{n}$ has an orthonormal basis $v_1,\dots, v_n$ where $A_1v_1 = \lambda v_1$ with $\lambda\neq 0$ and $A_1v_i = 0$ for $i\ge2$.

Let $A = A_1+A_2$ and $\|A_2\|=\delta$, then $$\| A v_1\| \le (\lambda + \delta)\text{, and }\| Av_i \| = \|A_2v_i\| \le \delta.$$

Recall that $|\det A|$ is the volume of the parallelepiped with edges $Av_i$, the statement follows.

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Edit:

I incorrectly claimed that $v_1,\dots, v_n$ can be chosen to be orthonormal. Instead, choose $v_1$ be a unit generator of $Im(A_1)$ and $v_2,\dots,v_n$ an orthonormal basis of $\ker (A_1)$. The difference is, here, $v_1$ needs not be perpendicular to $v_2,\dots, v_n$.

It is easy to show that $$\mathbb C^n = Im (A_1) \oplus \ker A_1,$$ i.e. $v_1,\dots, v_n$ is a basis of $\mathbb C^n$, noting again that $A_1 v_1 =\lambda v_1$.

The rest of the argument works, except that the last sentence should be change to:

$|\det A|$ is ratio of volumes of the parallelepiped with edges $A v_i$ and the parallelepiped with edges $v_i$, the later is a constant.