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Consider the following two sets of matrices:

  1. The first set is produced by $vv^T$ with $v = [v_1 \ \ v_2]^T$

$$\left\{\begin{bmatrix} v_1^2 & v_1v_2 \\ v_1v_2 & v_2^2 \end{bmatrix}: v_1^2+v_2^2 = 1\right\}$$

  1. 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.

Chill2Macht
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sleeve chen
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  • Do you mean are both sets equal or correspond bijectively? I.e. given any matrix in the first set, it equals a matrix in the second set, and vice versa? I'm not sure what you mean by isomorphism here, since the notion of equivalence has to be specified in advance for that term to be meaningful. – Chill2Macht Sep 11 '16 at 07:32
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    @William Actually two questions: 1. Are they equal? 2. Or are they isomorphic, which means there is a linear, one to one, and onto map between both sets and of course the inverse of such map is isomorphic. This map can be easily produced for example $f(x) = 1x$, where $x$ is the elements of one of each set, i.e. just multiply the element by one. – sleeve chen Sep 11 '16 at 07:35
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    $v_1^2+v_2^2=1$ $\Leftrightarrow$ $\ (v_1,v_2) \in \mathbb{U}$ (unit circle) $\Leftrightarrow \ v_1=\cos(\theta), v_2=\sin(\theta)$ for a certain $\theta$. That's all ! – Jean Marie Sep 11 '16 at 07:38
  • @JeanMarie Yes. However, actually I am writing a report and I want to give a relation between both sets. I have no idea what term I should use exactly and precisely. – sleeve chen Sep 11 '16 at 07:44
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    Then don't use the word isomorphic. Say plainly "are one and the same set by an evident bijection" – Jean Marie Sep 11 '16 at 07:53
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    (ctd) No normal mathematician will object to this word "evident" in this case. – Jean Marie Sep 11 '16 at 07:56

1 Answers1

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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.

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Chill2Macht
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  • Only one thing I am worried is that both sets are "infinite" sets. Can we say they are the same? My answer is they are same same though. – sleeve chen Sep 11 '16 at 07:47
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    I see what you are saying. One way to show that they are equal is the following. Let $A$ denote the first set and $B$ the second set. If one can show that, for any $a \in A$, that $a \in B$, i.e. that $a \in A \implies a \in B$, then one can conclude that $A \subset B$. Likewise, if for any $b \in B$, also $b \in A$, then $B \subset A$. If $A \subset B$ and $B \subset A$, then $A = B$. This is what Jean Marie's comment is saying – Chill2Macht Sep 11 '16 at 07:49
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    See e.g. p. 6 of this PDF http://www.people.vcu.edu/~rhammack/BookOfProof/SetProofs.pdf -- I remember seeing an answer on Math.SE that discussed the principle better, but I can't find it right now. It's really convenient because it works for any set, not just finite ones, but also infinite ones. – Chill2Macht Sep 11 '16 at 07:54