Consider the following two sets of matrices:
- The first set is produced by $vv^T$ with $v = [v_1 \ \ v_2]^T$
$$\left\{\begin{bmatrix} v_1^2 & v_1v_2 \\ v_1v_2 & v_2^2 \end{bmatrix}: v_1^2+v_2^2 = 1\right\}$$
- The second set is
$$\left\{\begin{bmatrix} \cos^2{\theta}& \cos{\theta}\sin{\theta} \\\cos{\theta}\sin{\theta} &\sin^2{\theta} \end{bmatrix}: 0\leq \theta \leq 2\pi\right\}$$
Obviously, both sets are infinite sets. Moreover, both sets consist of only positive semidefinite, rank one and trace one matrices.
Are both sets isomorphic? Or can we directly claim that they are the same sets?
Note: the restriction on the range of $\theta$ is to make both sets one to one.
These are the exact two same sets. I.e. they are equal, not just "isomorphic in some sense".
One way to see this is to note that for any $\theta$, $\cos^2 \theta + \sin^2 \theta =1$. Also note that "duplicate elements in a set are only counted once", i.e. the restriction on $\theta$ technically isn't necessary (I think), since if $\cos \theta_1 = \cos \theta_2$ and $\sin \theta_1 = \sin \theta_2$, the matrix corresponding to both $\theta_1$ and $\theta_2$ is still only considered to be in the set once.
In other words, your question becomes a lot easier when you remember that sets do not have repeated elements, only multisets do: Can elements in a set be duplicated?
The fact that the two sets are equal I think can also be made clearer if one considers the latter set to be of rotors in $\mathbb{R}^2$, which are 2-vectors/bivectors, then the first set just consists of the geometric products of "trigonometric" vectors which produce the 2-vectors corresponding to rotors -- although this argument probably is not helpful or useful since it requires background in geometric algebra, with which I myself am not familiar enough yet to be entirely sure that I am not making any technical mistake. In any case, the takeaway is that $v_1v_2^T$ corresponds to the tensor product, which also corresponds to the geometric product here, I believe.