Let $\rho(q):\mathbb{R}^3\rightarrow\mathbb{R}^3$ be a map defined by $q_1\mapsto qq_1q^{-1}$ for a unit quaternion $q$.
$S^3$ is the group of unit quaternions.
I know that $\rho(q)$ acts orthogonally (norm preserving linear map) on $\mathbb{R}^3$ by conjugation.
And, I know that $\rho:S^3 \to O(3)$ is a homomorhism. i.e. $\rho (ab)v = abvb^{-1}a^{-1}=a\rho(b)va^{-1}=\rho(a)(\rho(b)v)=\rho(a)\rho(b)v$ for any $v\in \mathbb{R}^3$, thus $\rho (ab)=\rho(a)\rho(b)$.
The proof (I will mark with dots where I need some help with):
$S^3$ is connected and as $\rho$ is continuous, the image of $\rho$ is connected.
The group $O(3)$ has two components: one ($SO(3)$) with matrices whose determinants are $1$, and the other component ($O(3) \setminus SO(3)$) has matrices whose determinants are $-1$.
- The component of the identity is $SO(3)$, so the image of $\rho$ is in $SO(3)$. (Why? Is it because $\rho$ is a group homomorphism so that the image must be a group?)
If $\rho(q)=I$, then $qq_1q^{-1}=q_1$ for all $q_1 \in \mathbb{R}^3$.
- We can see that this is only possible when $q \in \mathbb{R}$.
(I found that straight matrix multiplication of $qq_1q$ is pretty ugly, so I wonder how should this be shown. I tried applying a property of the trace, and determinant as well.)
The only reals in $S^3$ are $\pm1$, so the kernel of $\rho$ is $\{\pm 1\}$.
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It remains to prove that $\rho$ is onto. (Ok, so we've shown $\rho$ is well defined, and its kernel is $\{\pm1\}$. But, I wonder if it is sufficient to say $\rho$ is injective.)
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consider quaternion $q = \lambda + a \mu$ with $a \in \mathbb{R}^3$, $\left\|a \right\| =1$, $\lambda, \mu \in \mathbb{R}$ and $\lambda ^2 +\mu ^2 =1$. (I know that a quaternion must be in the form $q=x+iy+jz+ku$, and if it's a unit quaternion, then $x^2+y^2+z^2+u^2=1$. So, does $\lambda$ represent the real part, and $\mu$ represent the imaginary part?)
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Clearly, $qaq^{-1}=a$ so that $\rho(q)$ is a rotation with axis $a$. (I think this also means $qa=aq$, so it seems like they can commute. But, my textbook didn't really mention anywhere about the commutativity of quaternions, while it only says quaternions are very similar to complex numbers, which commute with $\mathbb{R}$. And, as they are similar to each other, I guess quaternions can commute with $\mathbb{R}^3$ by expanding $\{i\}$ to $\{i,j,k\}$. Is this correct?)
The trace of a rotation with angle $\theta$ is $1+2\cos \theta$, so to determine the angle of rotation, we compute the trace of $\rho(q)$.
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The diagonal entries of the matrix of $\rho(q)$ with repect to the basis $i,j,k$ are $\text{Re}(-iqiq), \text{Re}(-jqjq), \text{Re}(-kqkq)$. (Yes, the $\rho(q)$ is a rotation, so it should be in the form of $3 \times 3$ matrix. But, how do we compute diagonal entries?)
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If $a = a_1i+a_2j+a_3k$ and $q = \lambda + a \mu$, then $\text{Re}(-iqiq)= \lambda ^2 + \mu ^2(a_1^2-a_2^2-a_3^2)$. (I think understanding what $-iqiq, -jqjq, -kqkq$ are, this one and the last question would be solved. What do they really mean?)
Similarly for $j, k$.
So, $\text{tr}(\rho(q))=3 \lambda^2-\mu^2 \left\|a \right\| ^2=3 \lambda^2-\mu^2=1+ 2(\lambda^2 + \mu^2)=1+2\cos \theta $
This implies we can get any rotation by choosing $a,\lambda,\mu$ suitably. Thus, $\rho$ is onto.
Best Answer
The continuous image of a path-connected set is path-connected. Since the image contains the identity matrix, it can only contain matrices path-connected to the identity matrix, which is $SO(3)$.
I write quaternions as sums of scalars and 3D vectors. The product of two vectors has scalar and vector components given by the formula $\mathbf{uv}=-\mathbf{u}\cdot\mathbf{v}+\mathbf{u}\times\mathbf{v}$. You can use this to show (a) two quaternions commute iff their vector components are parallel and (b) two quaternions anticommute iff they are perpendicular vectors. In particular, a quaternion commutes with all vectors iff its vector component is parallel to all vectors iff its vector component is zero.
It's not sufficient. The first isomorphism theorem says a group mod a homomorphism's kernel is isomorphic to the image. Here, $\rho:S^3\to SO(3)$ has kernel $S^0$, so in order for the theorem to tell us $S^3/S^0\cong SO(3)$ we need to say the image is all of $SO(3)$.
The real part is $\lambda$, but the imaginary part is $a\mu$ not $\mu$. Sometimes I like to say "scalar" and "vector" parts (or components) instead of "real" and "imaginary." So the imaginary part is a 3D vector.
Yes, $qaq^{-1}=a$ and $qa=aq$ are equivalent in any group: just right-multiply by $q$ or $q^{-1}$ to convert one equation into the other.
Well, what does "quaternions can commute with R^3" mean, precisely? It's certainly not true that any quaternion commutes with all vectors. Like I said earlier, you can show two quaternions commute if and only if their vector components are parallel.
If $A$ is a matrix, the $ij$ entry is $e_i^TAe_j$, which is $\langle e_i,Ae_j\rangle$. In a general inner product space, to compute the $ij$ entry of a linear transformation with respect to a basis $\{e_i\}$, you compute $\langle e_i,Ae_j\rangle$.
For quaternions, the inner product is $\mathrm{Re}(\overline{x}y)=x_1y_1+x_2y_2+x_3y_3+x_4y_4$, and the basis for the subspace $\mathbb{R}^3$ is $\{\mathbf{i},\mathbf{j},\mathbf{k}\}$. Your formulas for the diagonal entries have typos: for instance the first diagonal entry is $\mathrm{Re}(-\mathbf{i}q\mathbf{i}\overline{q})$ (notice the conjugate, also $\overline{\mathbf{i}}=-\mathbf{i}$). So you would need to redo the calculations.
A smarter thing to do would be to pick a different basis that depends on the quaternion! Say you're talking about conjugation-by-$e^{\theta\mathbf{u}}$ (you should prove the sqrts of $-1$ are the unit vectors, Euler's formula holds for exponentials, and all quaternions have polar forms). Extend $\{\mathbf{u}\}$ to an orthonormal basis $\{\mathbf{u},\mathbf{v},\mathbf{w}\}$ and then compute the conjugates of these basis elements by $e^{\theta\mathbf{u}}$, then compute the length of their projections onto the original basis elements (i.e. the real parts of $\overline{\mathbf{x}}e^{\theta\mathbf{u}}\mathbf{x}e^{-\theta\mathbf{u}}$ for $\mathbf{x}\in\{\mathbf{i},\mathbf{j},\mathbf{k}\}$). This is not hard to do if you know parallel vectors commute, perpendicular vectors anticommute, unit vectors are sqrts of $-1$, and the product of two perpendicular vectors is a vector perpendicular to the first two.