TITLE: Topology of a set of matrices
QUESTION [5 upvotes]: I am trying to find the topology of $M$ where
$$M := \left\{ A \in \Bbb R^{3 \times 3} \mid A = 2 \vec{u} \cdot (\vec{u})^T - (\mathbb{1}_{3 \times 3} - \vec{u} \cdot (\vec{u})^T) \right\},$$
with $\vec{u}$ three dimensional unit vectors with real coefficients and $\mathbb{1}_{3 \times 3}$ the $3 \times 3$ identity matrix.
What I have done so far is putting $\vec{u} = (a \ \ b \ \ c)^T$, with $a^2 + b^2 + c^2 = 1$, from which $A$ takes the general form
\begin{align*}
A = \begin{pmatrix} 3a^2-1 & 3ab & 3ac \\ 3ab & 3b^2 - 1 & 3bc \\ 3ac & 3bc & 3c^2 -1 \end{pmatrix}.
\end{align*}
From this general form I deduced that $A$ has determinant 2 (and is therefore invertible), $A$ is symmetric, and has a vanishing trace. At this point I do not quite see what the topology of $M$ would be, however, I should mention that I am not very familiar with topology yet.
Any help would be greatly appreciated.

REPLY [3 votes]: The  map $f$ from the unit sphere in $\mathbb{R}^3$ to your space is continuous and $f(-\vec{u}) = f(u).$ So, your space is diffeomorphic to the projective plane $\mathbb{R}P^2.$