Necessary and sufficient conditions for rank-1 connection My question concerns the following problem: 
Given a positive diagonal matrix $\alpha$ with $\det\alpha\geq 1$, when is it possible to find non-zero vectors $a,n\in\mathbb{R}^n$ with $|n|=1$ and an orthogonal matrix $Q$ with $\det Q = 1$ such that $Q\alpha = I + an^T$?
For example, if $\alpha=I$ then this is impossible. Proof: suppose that $Q= I + an^T$. Let $P$ be an orthogonal matrix with $\det P = 1$ whose last column is $n$. Then $QP$ is an orthogonal matrix, whose first $n-1$ columns equal those of $P$. Since $\det P = 1$, the last column of $QP$ therefore equals $n$. But if $a\neq 0$ then $Qn = a + n \neq n$.
There are intriguing results I already know, but I don't like the proof in 2D and I don't know the proof in 3D. Also, I like the problem so therefore I share it here.
In 2D, if $$\alpha = \pmatrix{\alpha_1 & 0 \\ 0 & \alpha_2},\,\alpha_1\alpha_2\geq 1, 0<\alpha_1\leq\alpha_2$$
then one can find orthogonal $Q$ and vectors $a,n$ such that $Q = I + an^T$ if and only if $\alpha_1\leq 1$. In 3D, if
$$\alpha = \pmatrix{\alpha_1 & 0 & 0 \\ 0 & \alpha_2 & 0\\0 & 0 & \alpha_3},\,\alpha_1\alpha_2\alpha_3\geq 1, 0<\alpha_1\leq\alpha_2\leq \alpha_3$$ one can do it if and only if $\alpha_2 = 1$.
Does anyone know the necessary and sufficient conditions on $\alpha$ in $n$ dimensions?
 A: Let $A\in M_n(\mathbb R)$ be a positive diagonal matrix with $\det(A)\ge1$. The equation
$$
QA-I = uv^T\tag{1}
$$
subject to $Q\in SO(n,\mathbb R),\,u,v\in\mathbb R^n$ and $\|u\|_2=1$ implies that
$$
A^2-I = (QA)^T(QA)-I = uv^T+vu^T+vv^T.\tag{2}
$$
Since the LHS is a diagonal matrix and the rank of the RHS is at most two, equations $(1)$ and $(2)$ are solvable only if $A^2-I$ has at most two nonzero diagonal entries. It follows that up to a permutation of the diagonal entries, $A$ must be in the form of
$$
A=\operatorname{diag}(a,b,1,1,\ldots,1)
\text{ where } 0<a\le b \text{ and } ab=\det(A)\ge1.\tag{3}
$$
Equation $(1)$ also implies that $QA=I+uv^T$. Therefore,
\begin{cases}
ab=\det(A)=\det(QA)=\det(I+uv^T)=1+v^Tu,\\
a+b+n-2=\operatorname{tr}(A)\ge\operatorname{tr}(QA)=\operatorname{tr}(I+uv^T)=n+v^Tu.
\end{cases}
It follows that $a+b+n-2\ge n+(ab-1)$, i.e.
$$
a+b\ge ab+1
\ \Leftrightarrow\ (a-1)(b-1)\le0
\ \Leftrightarrow\ a\le1\le b.\tag{4}
$$
Conditions $(3)$ and $(4)$ are necessary for $(1)$ to be solvable. They also comprise a sufficient condition. Suppose that $A$ takes the form of $(3)$ and $(4)$. Let
$$
B=\pmatrix{\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta}\pmatrix{a\\ &b}-I_2
=\pmatrix{a\cos\theta-1&-b\sin\theta\\ a\sin\theta&b\cos\theta-1}.
$$
Since $0<ab+1\le a+b$,
$$
\det(B)=ab+1-(a+b)\cos\theta
$$
can be made zero by choosing an appropriate $\theta$. With this $\theta$, $B$ is singular. Hence $B=xy^T$ for some unit vector $x\in\mathbb R^2$ and some vector $y\in\mathbb R^2$. It follows that $(1)$ is solvable by taking
$$
Q=\pmatrix{\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\\ &&I_{n-2}},
\ u=\pmatrix{x\\ \mathbf0_{n-2}},
\ v=\pmatrix{y\\ \mathbf0_{n-2}},
$$
where $\mathbf0_{n-2}$ denotes the zero vector with $n-2$ elements. 
