Consider the map $f : S^{m} \rightarrow Sym(\Bbb R^{m+1})$ from the unit sphere into the space of symmetric matrices that takes a point to the reflection about the corresponding line. It is claimed that we can "identify" the real projective space $\mathbb{R}P^{m}$ with the image $f(S^{m}$).
What does "identify" mean in this context? $\mathbb{R}P^{m}$ is the space of equivalence classes of lines in $\mathbb{R^{m+1}}$ and $f(S^{m})$ is a subset of matrices. The only thing they have incommon is that two points are sent to one in the same fashion, is that all there is to it?
There are two possible viewpoints for the real projective space of dimension $m$. You can see it as the quotient space obtained from $\Bbb R^{m+1}-0$ by identifying collinear points (i.e it's the space of lines in $\Bbb R^{m+1}$). You can see it at the quotient obtained from the $m$-sphere $S^m$ by identifying antipodal points. See here.
If $x,y\in S^m$ then you can check that $f(x)=f(y)$ if and only if $x=y$ or $x=-y$. Take the viewpoint of the real projective space as a quotient of the $m$-sphere. The continuous map $f$ factors to a continous bijection $$ \widetilde{f}:\Bbb RP^m\longrightarrow f(S^m).$$ Because the projective space is compact and $f(S^m$) is Hausdorff, $\widetilde{f}$ is a homeomorphism. This is how you can identify the projective space and $f(S^m)$.