Measuring 3D Rotation from Two Values

Suppose there is an matrix $T$ with unknown elements, but it it known that $T$ is a linear transformation representing a series of rotations.

Without knowing anything about how this rotation was performed (there are a number of ways to represent it anyway), how much information can I obtain about this rotation from only the following information:

The projection of the transformed x-axis onto the original x-axis; that is: $$\begin{bmatrix}1 \\ 0\\ 0\end{bmatrix}\cdot T\begin{bmatrix}1 \\ 0\\ 0\end{bmatrix}$$ And the projection of the transformed z-axis onto the original z-axis, that is: $$\begin{bmatrix}0 \\ 0\\ 1\end{bmatrix}\cdot T\begin{bmatrix}0 \\ 0\\ 1\end{bmatrix}$$

From these "measurements", is it possible to determine enough information such that there is only one degree of freedom in constructing $T$ from a series of rotations?

• What do you mean by “matrix representing some random orientation?” What exactly is in this matrix? – amd Sep 1 '17 at 0:45
• @amd it's a 3x3 matrix representing the transformation from the original coordinate system to the rotated one. This matrix could be the product of multiple 3D rotation matrices, for instance. – Billy Kalfus Sep 1 '17 at 0:48
• Then you already have the transformed axes in their entirety. They’re the columns of the matrix. – amd Sep 1 '17 at 1:46
• Yes - but the problem is I don't know this matrix. I only have two elements of it. – Billy Kalfus Sep 1 '17 at 16:44
• That’s not what you wrote in your question, though: “Suppose I have a matrix...” – amd Sep 1 '17 at 18:40

In a sense, yes. For instance, in the Wikipedia page you linked to, there is a way to represent a rotation by a unit quaternion $q=w+ix+jy+kz$ (with $w^2+x^2+y^2+z^2=1$ and $w\ge0$): $$T = \begin{bmatrix} 1 - 2 y^2 - 2 z^2 & 2 x y - 2 z w & 2 x z + 2 y w \\ 2 x y + 2 z w & 1 - 2 x^2 - 2 z^2 & 2 y z - 2 x w \\ 2 x z - 2 y w & 2 y z + 2 x w & 1 - 2 x^2 - 2 y^2 \end{bmatrix}.$$ So, if $t_{11}$ and $t_{33}$ are given, define $a=\frac12(1-t_{11})$ and $b=\frac12(1-t_{33})$. Then $y^2+z^2=a$ and $x^2+y^2=b$. Hence $$\sqrt{\max\left(\min(a,\,b),\,a+b-1\right)}\le|y|\le\sqrt{\max(a,\,b)}.$$ Once you have chosen the free variable $|y|$ in the above range, you can determine $|x|,|z|$ and $w$ uniquely according to $x^2+y^2=b,\ y^2+z^2=a$ and $w^2+x^2+y^2+z^2=1$. Picking also the signs for $x,y,z$, you get eight different rotation matrices for each feasible $|y|$. In other words, there are eight loci of $T$, on each there is only one degree of freedom.
Each projection determines a circle on which the transformed unit vector must lie. For the transformed $x$-axis $\mathbf u$, we know $u_x$, so this circle is $x=u_x$, $y^2+z^2=1-u_x^2$. Similarly, the transformed $z$-axis, $\mathbf w$, lies somewhere on $z=w_z$, $x^2+y^2=1-w_z^2$. For any choice of $\mathbf u$, the corresponding $\mathbf w$ lies on the intersection of the latter circle and the plane $\mathbf u\cdot\mathbf p=0$, hence there can be zero, one or two corresponding vectors $\mathbf w$. These two vectors are the first and third columns of the rotation matrix. There are no intersections when $u_z^2\gt1-w_z^2$, so this constrains the possibilities for $\mathbf u$ somewhat. A symmetric constraint comes from considering the intersection of the plane $\mathbf w\cdot\mathbf p=0$ with the first circle: $w_x^2\le1-u_x^2$. The transformed $y$-axis, and hence the middle column of the rotation matrix is, of course, completely determined by $\mathbf u$ and $\mathbf w$.
• @BillyKalfus No. Here, $\mathbf p$ is a variable: every point $\mathbf p$ on that plane satisfies the equation. You right, though, that in particular we want $\mathbf u\cdot\mathbf w=0$ as well. – amd Sep 3 '17 at 18:43