欧拉四平方和恒等式与矩阵

青青子衿

\begin{align*}   
\boldsymbol{A}\left[m_{\overset{\,}1},m_{\overset{\,}2},m_{\overset{\,}3},m_{\overset{\,}4}\right]:=\underset{\circledcirc\,\!m}{\boldsymbol{A}}&=\begin{pmatrix}   
m_{\overset{\,}1}&m_{\overset{\,}2}&m_{\overset{\,}3}&m_{\overset{\,}4}\\   
-m_{\overset{\,}2}&m_{\overset{\,}1}&-m_{\overset{\,}4}&m_{\overset{\,}3}\\   
m_{\overset{\,}3}&-m_{\overset{\,}4}&-m_{\overset{\,}1}&m_{\overset{\,}2}\\   
-m_{\overset{\,}4}&-m_{\overset{\,}3}&m_{\overset{\,}2}&m_{\overset{\,}1}\\   
\end{pmatrix}\\   
\\
\boldsymbol{A}\left[n_{\overset{\,}1},n_{\overset{\,}2},n_{\overset{\,}3},n_{\overset{\,}4}\right]:=\underset{\circledcirc\,\!n}{\boldsymbol{A}}&=\begin{pmatrix}   
n_{\overset{\,}1}&n_{\overset{\,}2}&n_{\overset{\,}3}&n_{\overset{\,}4}\\   
-n_{\overset{\,}2}&n_{\overset{\,}1}&-n_{\overset{\,}4}&n_{\overset{\,}3}\\   
n_{\overset{\,}3}&-n_{\overset{\,}4}&-n_{\overset{\,}1}&n_{\overset{\,}2}\\   
-n_{\overset{\,}4}&-n_{\overset{\,}3}&n_{\overset{\,}2}&n_{\overset{\,}1}\\   
\end{pmatrix}\\   
\\   
\underset{\circledcirc\,\!m}{\boldsymbol{A}}^{\rm{T}}\underset{\circledcirc\,\!m}{\boldsymbol{A}}   
&=({m_{\overset{\,}1}}\!^2+{m_{\overset{\,}2}}\!^2+{m_{\overset{\,}3}}\!^2+{m_{\overset{\,}4}}\!^2)\,\boldsymbol{I}\\
\underset{\circledcirc\,\!n}{\boldsymbol{A}}^{\rm{T}}\underset{\circledcirc\,\!n}{\boldsymbol{A}}   
&=({n_{\overset{\,}1}}\!^2+{n_{\overset{\,}2}}\!^2+{n_{\overset{\,}3}}\!^2+{n_{\overset{\,}4}}\!^2)\,\boldsymbol{I}\\
\big(\underset{\circledcirc\,\!m}{\boldsymbol{A}}\,\underset{\circledcirc\,\!n}{\boldsymbol{A}}\big)\!^{\rm{T}}\big(\underset{\circledcirc\,\!m}{\boldsymbol{A}}\,\underset{\circledcirc\,\!n}{\boldsymbol{A}}\big)&=\underset{\circledcirc\,\!s}{\boldsymbol{A}}^{\rm{T}}\underset{\circledcirc\,\!s}{\boldsymbol{A}}\\
&=({s_{\overset{\,}1}}\!^2+{s_{\overset{\,}2}}\!^2+{s_{\overset{\,}3}}\!^2+{s_{\overset{\,}4}}\!^2)\,\boldsymbol{I}
\end{align*}

rpi.edu/dept/arc/training/latex/LaTeX_symbols.pdf


\begin{align*}
&\left\{\begin{split}
y^2&=\alpha ^2+b x+c x^2+d x^3+e x^4\\
\\
x&=\frac{8 \alpha ^2 \left(b \left(4 \alpha ^2 c-b^2\right)-8 \alpha ^4 d\right)}{64 \alpha ^6 e-\left(4 \alpha ^2 c-b^2\right)^2}\\
y&=\alpha+\frac{b}{2 \alpha}x+\frac{4\alpha ^2 c-b^2}{8\alpha ^3}x^2\\
\end{split}\right.\\
\\
&\left\{\begin{split}
\eta^2&=a+b\xi+c\xi^2+d\xi^3+\varepsilon ^2\xi^4\\
\\
\xi&=\frac{64\varepsilon^6a-\left(4\varepsilon^2c -d^2\right)^2}{8\varepsilon ^2 \left(d \left(4\varepsilon ^2c-d^2\right)-8\varepsilon ^4b \right)}\\
\eta&=\frac{4\varepsilon ^2c-d^2}{8\varepsilon ^3}+\frac{d}{2\varepsilon }\xi+\varepsilon\xi^2\\
\end{split}\right.\\
\\
&\left\{\begin{split}
Y^2&=a+bX+cX^2+dX^3+eX^4\\
q^2&=a+bp+cp^2+dp^3+ep^4\\
\\
X&=p+\frac{8 q^2 \left(\mathcal{D} \left(4 \mathcal{C}-\mathcal{D}^2\right)-8 \mathcal{B}\right)}{64 \mathcal{A}-\left(4 \mathcal{C}-\mathcal{D}^2\right)^2}\\
Y&=q+\frac{\mathcal{D} (X-p)}{2 q}+\frac{\left(4 \mathcal{C}-\mathcal{D}^2\right) (X-p)^2}{8 q^3}\\
\mathcal{A}&=e q^6\\
\mathcal{B}&=(d+4 e p)q^4\\
\mathcal{C}&=\left(c+3 d p+6 e p^2\right)q^2\\
\mathcal{D}&=b+2 c p+3 d p^2+4 e p^3
\end{split}\right.
\end{align*}

hbghlyj

MSE
The Quaternions $\mathbb{H}$ are isomorphic (as $\mathbb{R}$-algebras) to the given set of matrices. The isomorphism looks like this:
$$
\phi: a + bi + cj + dk \longmapsto \begin{pmatrix}a & b & c & d  \\  -b & a & -d & c \\ -c &d &a& - b\\ -d& -c & b& a\end{pmatrix}.
$$
To "understand" why this is true, you "simply" check that this is an isomorphism.

You check for example that $\phi$ is bijective, which is clear from the construction.

Then you check that $\phi$ is an algebra homomorphism, so you need for $x,y\in \mathbb{H}$ and $\lambda \in \mathbb{R}$:

• $\phi(xy) = \phi(x)\phi(y)$ for $x,y\in\mathbb{H}$
• $\phi(x+y) = \phi(x) + \phi(y)$
• $\phi(\lambda x) = \lambda\phi(x)$

The last two are not difficult to check. The first one requires a bit of work.

Even though this does not answer the minus signs are where there are in the matrix, I highly recommend that you try to prove that $\phi$ is a homomorphism. This exercise will make you more familiar with the Quaternions.

But note if you check property $3$ above you would need (as a special case)
$$
\phi((bi)(bi))  = \phi(ib)\phi(ib).
$$
That is you would need
$$
\begin{pmatrix} -b^2 & 0 & 0& 0  \\  0 & -b^2 & 0 & 0 \\ 0 &0 &-b^2& 0\\ 0& 0 & 0& -b^2\end{pmatrix} = \begin{pmatrix} 0 & b & 0& 0  \\  -b & 0 & 0 & 0 \\ 0 &0 &0& -b\\ 0& 0 & b& 0\end{pmatrix}\begin{pmatrix} 0 & b & 0& 0  \\  -b & 0 & 0 & 0 \\ 0 &0 &0& -b\\ 0& 0 & b& 0\end{pmatrix}.
$$
So here you can see that you "need" the minus on all the $b$'s. In this case it comes down to the fact that $i^2 = -1$.