Given $v\in\mathbb R^3$, what are the solutions $A\in\mathbb R^{3\times 3}$ of $A^2=I+vv^{T}$? The following question arose while deriving the explicit form of symmetric Lorentz transformations:

Given $v\in\mathbb R^3$, what can we say about the set
\begin{equation}
\{A\in\mathbb R^{3\times 3}:A^2=I+vv^{T}\}\ 
\end{equation}
(with $I\in\mathbb R^{3\times 3}$ being the identity matrix)?

Are there any analogies to the case where $A$ and $v$ are real numbers, i.e. are there exactly two solutions which differ by a minus sign?
 A: If $a,b\in\mathbb R^3$ are unit vectors such that $\{ v/\|v\|,a,b \}$ is an orthonormal base of $\mathbb R^3$, then any $A$ of the form $\pm\frac{\sqrt{1+\|v\|^2}}{\|v\|^2}vv^T\pm aa^T\pm bb^T$ will work. Conversely if $A$ is a solution to your equation then it can be written like I described, the reason is that the square root of a positive definite symmetric matrix $U \Sigma U^T$ are $U\begin{bmatrix} \pm\sqrt{\Sigma_{11}}&0&0\\0&\pm\sqrt{\Sigma_{22}}&0\\0&0&\pm\sqrt{\Sigma_{33}} \end{bmatrix} U^T$ and your equation forces one eigen vector of $I+vv^T$ to be $v$, the rest is free (two dimensions of $U$ and the $\pm$ signs).
A: In this answer, we don't assume that $A$ is necessarily symmetric.
Suppose that $v\not = 0$ and let $w$ be any vector independent from $v$. Suppose that $A v = w$. We have $A w = A^2 v = (1 + |v|^2) v$,
hence $A^2 w = A(A w) = (1 +|v|^2) A v= (1 +|v|^2) w$, but by hypothesis, $A^2 w = w + (v^T w) v$ hence $|v|^2 w  = (v^T w) v$, a contradiction.
It follows that $v$ and $Av$ are necessarily colinear, hence $A v = r v$ for some number $r$, hence $A^2 v = r A v = r^2 v$ but $A^2 v = (1 + |v|^2) v $ hence $r = \pm\sqrt{1 + |v|^2}$ hence
\begin{equation}
A v = \pm \sqrt{1 + |v|^2} v := r v
\end{equation}
Now we have
\begin{equation}
A^3 = A + (Av)v^T = A + r v v^T = A + r (A^2 - I)
\end{equation}
hence $A$ annihilates the polynomial
\begin{equation}
P(x) = x^3 - r x^2 - x + r = (x^2-1)(x - r)
\end{equation}
It follows that besides $r$, the only possible complex eigenvalues of $A$ are $1$ and $-1$, furthermore, any eigenvector not colinear to $v$ is also an eigenvector of $A^2$, hence it must be orthogonal to $v$.
The minimal polynomial cannot be $x-r$ because $A\not = r I$, hence there is at least one non zero eigenvector $w\in v^\perp$ such that $A w = \epsilon w$ with $\epsilon=\pm 1$. Let now $u$ be a vector in $v^\perp$ independent from $w$.
Suppose that $A u = \alpha v + \beta w + \gamma u$, then
\begin{equation}
u = A^2 u = \alpha r v + \beta \epsilon w + \gamma
(\alpha v + \beta w + \gamma u)
\end{equation}
This implies
\begin{equation}
\gamma^2 = 1,\quad\alpha(\gamma+r) = 0,\quad\beta(\gamma +\epsilon)=0
\end{equation}
As $\gamma + r = r \pm 1 \not = 0$, it follows that $\alpha = 0$ which means that the plane ${\cal P} = v^\perp$ is stable by $A$.
The restriction $A_P$ of $A$ to ${\cal P}$ satisfies $A_P^2 = I_P$ which means that $A_P$ is either $\pm I_P$ or a (non necessarily orthogonal) symmetry with respect to a line, that is to say it has an eigendirection for the eigenvalue +1 and an eigendirection for the eigenvalue -1.
Such a matrix can be represented as
\begin{equation}
A = \pm\frac{\sqrt{1 + |v|^2}}{|v|^2}v v^T
\pm \frac{w (v\times u)^T}{\det(v, w, u)}
\pm \frac{u (v\times w)^T}{\det(v, w, u)}
\end{equation}
where $w$ and $u$ are independent vectors orthogonal to $v$ but not necessarily mutually orthogonal.
A: When $v\neq 0$ then go to an o.n.b $(e_1,e_2,e_3)$ with say $e_3$ in the direction of $v$. Then $C=I+vv^T$ takes the diagonal form $C={\rm diag}(1,1,\lambda^2)$ with $\lambda>1$. If $f_3=Ae_3$ was linearly independent from $e_3$ then $A f_3=\lambda^2 e_3$ would imply that the kernel of $A^2-\lambda^2$ was two dimensional which is not the case. Thus $Ae_3=\pm \lambda e_3$. $A$ has to preserve the the span of the first two basis vectors $V=\langle e_1,e_2\rangle$ and $A^2$ is the identity on $V$. When $A$ is symmetric this implies that $A$ is the identity, minus the identity (on $V$)
or a reflection. For the latter case you have:
$$ A = \pmatrix{a & b & 0 \\ b & -a &0 \\ 0 & 0 & \pm\lambda}$$
for any $a,b$ with $a^2+b^2=1$ (the eigenvalues are thus $1,-1,\pm\lambda$).
When $v=0$, $A^2=I$. Then $P=(I+A)/2$ and $Q=1-P=(I-A)/2$ are orthogonal projections and $AP=P$, $AQ=-Q$. If either of them has zero rank $A=\pm I$. If $P$ has rank 2 you again have a reflection $A =I-2nn^T$ for some unit vector $n$, and if $Q$ has rank 2, then $A$ is minus a reflection $A=2nn^T -I$. I think this goes through all the possible symmetric solutions.
The non-symmetrical ones yield
a somewhat larger space of solutions.
A: When v=0 we get $A^2=I$ and hence we get A. When v$\ne$0 then $vv^T$ is a rank one positive semi definite matrix and hence I+$vv^T$ is a positive definite matrix with a unique square root.
