Given vectors $u,v\in \mathbb{R}^d$, I wonder what we can say about the minimum singular value of $I-uv^\top$? I know that when $u=v$, this matrix is symmetric so it is not hard to compute this. Specifically, I am consider the case where $u=\mathbf{1}$ and $v=\frac{x}{1+\mathbf{1}^\top x}$. Thanks!
== Update: Actually this matrix has (n-1) linearly-independent eigenvectors $c$'s that are perpendicular to $x$ (with eigenvalue 1) and eigenvector $\mathbf{1}$ with eigenvalue $\frac{1}{1+1^\top x}$. Suppose that $1^\top x > 0$ (which is the case). Then we actually know all the eigenvalues/vectors for this $A=I-uv^\top$ matrix but I do not know how this relates to the singular value of $A$.
== Update 2: It would be also very helpful if you know something about singular value of $u^\top v\cdot I - uv^\top$ with the above form of $u=\mathbf{1}$ and $v$ with all-positive entries.
Let $A=I-uv^\top$ and $V=\operatorname{span}\{u,v\}$. Since $AV\subseteq V$ and $A|_{V^\perp}=\operatorname{id}$, at least $n-2$ singular values of $A$ are equal to $1$. Denote the remaining two singular values by $s_1$ and $s_2$. Then $$ s_1s_2=|\det A|=|1-v^\top u| $$ and $$ \begin{aligned} s_1^2+s_2^2 &=\|A\|_F^2-(n-2)\\ &=\operatorname{tr}\big((I-uv^\top)(I-vu^\top)\big)-(n-2)\\ &=2(1-v^\top u)+\|u\|^2\|v\|^2. \end{aligned} $$ Therefore $$ \begin{aligned} s&:=\min\{s_1,s_2\}\\ &=\frac{\sqrt{s_1^2+s_2^2+2s_1s_2}-\sqrt{s_1^2+s_2^2-2s_1s_2}}{2}\\ &=\frac{\sqrt{\|u\|^2\|v\|^2+2(1-v^\top u)+2|1-v^\top u|}\,-\,\sqrt{\|u\|^2\|v\|^2+2(1-v^\top u)-2|1-v^\top u|}}{2}\\ &=\frac{\sqrt{\|u\|^2\|v\|^2+4\max\{1-v^\top u,\,0\}}\,-\,\sqrt{\|u\|^2\|v\|^2+4\min\{1-v^\top u,\,0\}}}{2}.\\ \end{aligned} $$ It may not seem obvious, but the second largest singular value of $A$ is actually always $\le1$. Hence the minimum singular value of $A$ is $s$.