Alfredeim

2022-09-15

Given the parameters ${p}_{1},{p}_{2},{p}_{3}$ does the following system can be solved:

This system actually discribes a Rotation Matrix

${p}_{1}\text{}\text{}a\text{}\text{}\text{}\text{}b\phantom{\rule{0ex}{0ex}}c\text{}\text{}\text{}\text{}{p}_{2}\text{}\text{}d\phantom{\rule{0ex}{0ex}}e\text{}\text{}\text{}\text{}f\text{}\text{}\text{}\text{}{p}_{3}$

where only ${p}_{1}$, ${p}_{2}$ and ${p}_{3}$ are know. For example ${p}_{1}={p}_{2}={p}_{3}=1$ would result in $a=b=c=d=e=f=0$.

Why is this system (unabigiously) solvable / not solvable?

If it is solvable, what is the solution?

This system actually discribes a Rotation Matrix

${p}_{1}\text{}\text{}a\text{}\text{}\text{}\text{}b\phantom{\rule{0ex}{0ex}}c\text{}\text{}\text{}\text{}{p}_{2}\text{}\text{}d\phantom{\rule{0ex}{0ex}}e\text{}\text{}\text{}\text{}f\text{}\text{}\text{}\text{}{p}_{3}$

where only ${p}_{1}$, ${p}_{2}$ and ${p}_{3}$ are know. For example ${p}_{1}={p}_{2}={p}_{3}=1$ would result in $a=b=c=d=e=f=0$.

Why is this system (unabigiously) solvable / not solvable?

If it is solvable, what is the solution?

Dalton Erickson

Beginner2022-09-16Added 10 answers

Let

$$M=\left(\begin{array}{ccc}{p}_{1}& a& b\\ c& {p}_{2}& d\\ e& f& {p}_{3}\end{array}\right).$$

If $M$ is a rotation matrix, then ${M}^{-1}={M}^{T}.$ This implies that $M$ and ${M}^{-1}$ have the same entries on the main diagonal. But if $M$ is not the identity, $M\ne {M}^{-1},$ so the rotation matrix is not completely determined by the entries on its main diagonal.

On the other hand, if $[{u}_{1}\text{}{u}_{2}\text{}{u}_{3}{]}^{T}$ is a unit vector on the axis of rotation of $M$ and if 𝜃 is the angle of rotation about that axis, then

$$\begin{array}{}\text{(1)}& {p}_{i}=(1-\mathrm{cos}\theta ){u}_{i}^{2}+\mathrm{cos}\theta \end{array}$$

for $i=1,2,3.$ Moreover,

$${p}_{1}+{p}_{2}+{p}_{3}=1+2\mathrm{cos}\theta .$$

Therefore we can express cos𝜃 in terms of ${p}_{1}.$, ${p}_{2}.$, and ${p}_{3}.$. Plug that value of cos𝜃 into equation (1) for each $i;$ this either gives ${u}_{1}=0$ or gives two possible values of ${u}_{1}$ which differ only by a sign change.

We can safely assume that $0\le \theta \le \pi ,$, because the rotation described by angle $-\theta $ and unit vector $[{u}_{1}\text{}{u}_{2}\text{}{u}_{3}{]}^{T}$ is the same as the rotation described by angle $\theta $ and unit vector $[-{u}_{1}\text{}-{u}_{2}\text{}-{u}_{3}{]}^{T}.$. That means that in general there are eight possible ways to fill in the matrix $M$ (one for each choice of the signs of each of the ${u}_{i}),$), therefore eight possible solutions to the given set of equations. (For $0<\theta <\pi ,$, there are four solutions if exactly one of the ${u}_{i}$ is zero, two solutions if two of the 𝑢𝑖 are equal to zero. There are half as many solutions if $\theta =\pi ,$, and of course only one solution if $\theta =0.)$.) Moreover, by computing the rotation matrix for the rotation by angle $\theta $ around the axis given by $[{u}_{1}\text{}{u}_{2}\text{}{u}_{3}{]}^{T},$, we can compute all the unknown entries $a,b,c,d,e,$ and $f$ in the rotation matrix for a specific choice of $[{u}_{1}\text{}{u}_{2}\text{}{u}_{3}{]}^{T}.$

$$M=\left(\begin{array}{ccc}{p}_{1}& a& b\\ c& {p}_{2}& d\\ e& f& {p}_{3}\end{array}\right).$$

If $M$ is a rotation matrix, then ${M}^{-1}={M}^{T}.$ This implies that $M$ and ${M}^{-1}$ have the same entries on the main diagonal. But if $M$ is not the identity, $M\ne {M}^{-1},$ so the rotation matrix is not completely determined by the entries on its main diagonal.

On the other hand, if $[{u}_{1}\text{}{u}_{2}\text{}{u}_{3}{]}^{T}$ is a unit vector on the axis of rotation of $M$ and if 𝜃 is the angle of rotation about that axis, then

$$\begin{array}{}\text{(1)}& {p}_{i}=(1-\mathrm{cos}\theta ){u}_{i}^{2}+\mathrm{cos}\theta \end{array}$$

for $i=1,2,3.$ Moreover,

$${p}_{1}+{p}_{2}+{p}_{3}=1+2\mathrm{cos}\theta .$$

Therefore we can express cos𝜃 in terms of ${p}_{1}.$, ${p}_{2}.$, and ${p}_{3}.$. Plug that value of cos𝜃 into equation (1) for each $i;$ this either gives ${u}_{1}=0$ or gives two possible values of ${u}_{1}$ which differ only by a sign change.

We can safely assume that $0\le \theta \le \pi ,$, because the rotation described by angle $-\theta $ and unit vector $[{u}_{1}\text{}{u}_{2}\text{}{u}_{3}{]}^{T}$ is the same as the rotation described by angle $\theta $ and unit vector $[-{u}_{1}\text{}-{u}_{2}\text{}-{u}_{3}{]}^{T}.$. That means that in general there are eight possible ways to fill in the matrix $M$ (one for each choice of the signs of each of the ${u}_{i}),$), therefore eight possible solutions to the given set of equations. (For $0<\theta <\pi ,$, there are four solutions if exactly one of the ${u}_{i}$ is zero, two solutions if two of the 𝑢𝑖 are equal to zero. There are half as many solutions if $\theta =\pi ,$, and of course only one solution if $\theta =0.)$.) Moreover, by computing the rotation matrix for the rotation by angle $\theta $ around the axis given by $[{u}_{1}\text{}{u}_{2}\text{}{u}_{3}{]}^{T},$, we can compute all the unknown entries $a,b,c,d,e,$ and $f$ in the rotation matrix for a specific choice of $[{u}_{1}\text{}{u}_{2}\text{}{u}_{3}{]}^{T}.$

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