Question bank — DCM kinematics — Ċ = −[ω×]C

True — the equation is linear in $C$: the derivative is $C$ left-multiplied by a fixed (at that instant) skew matrix. That is why the solution is a matrix exponential, not a polynomial.

False — $\dot{C}$ lives in the tangent space to SO(3), not on the group. It is (skew) $\times$ (rotation), which is generally neither orthogonal nor of determinant 1.

True — zero spin gives $\dot{C}=-[0\times ]C=0$, so $C$ freezes. Physically the body is not turning, so the map between frames cannot change.

True — it follows directly from differentiating the constraint $C{C}^T=I$, which holds for all $t$. That single fact is what forces the whole skew structure.

False for the exact ODE — the equation guarantees $\frac{d}{dt}(C{C}^T)=0$, so orthonormality is conserved automatically once true initially. It is only numerical integration that must re-impose it.

True — for the inverse map you get ${\dot{C}}^T=+[{\vec{\omega }}_{\mathcal{B}}\times ]C^T$. The sign is a bookkeeping consequence of which direction the matrix maps, not a physical fact.

True — a $3\times 3$ skew matrix has zeros on the diagonal and one sign-flipped copy of each off-diagonal, leaving 3 free entries. These map one-to-one to $({\omega }_1,{\omega }_2,{\omega }_3)$ via the vee-map.

False in general — the clean exponential holds only when $[{\vec{\omega }}_{\mathcal{B}}\times ]$ is constant. If $\vec{\omega }$ varies in time and the skew matrices at different times don't commute, you need a time-ordered exponential, not $e^{\int (\cdot )dt}$.

Wrong frame — with $C:\mathcal{N}\to \mathcal{B}$ the skew must use body-frame components ${\vec{\omega }}_{\mathcal{B}}=C{\vec{\omega }}_{\mathcal{N}}$, so the output lands consistently in the body frame. Using ${\vec{\omega }}_{\mathcal{N}}$ mismatches the frames on the two sides.

The plus formula describes a vector seen from the inertial frame; $C$ maps inertial into body, so from the body's viewpoint inertial vectors appear to spin backward — hence the minus. The two formulas describe opposite viewpoints.

The continuous ODE preserves $C{C}^T=I$ exactly; a discrete Euler step takes a straight line off the curved surface SO(3), so $C$ drifts. You must renormalize (Gram–Schmidt / SVD) or propagate a quaternion.

Sign error — since $[{\vec{\omega }}_{\mathcal{B}}\times ]=-\dot{C}{C}^T$, the correct extraction carries a minus: ${\omega }_i=-\frac{1}{2}{\epsilon }_{ijk}(\dot{C}{C}^T{)}_{jk}$. Dropping it flips the recovered spin direction.

Only for proper rotations ($\det C=+1$). A reflection ($\det C=-1$) flips handedness and introduces an extra minus sign, breaking the identity used in the sign derivation.

A $3\times 3$ skew matrix is singular (odd-dimensional skew matrices have determinant 0), so it has no inverse. Recovery uses the transpose trick $[\omega \times ]=-\dot{C}{C}^T$, exploiting $C^{-1}=C^T$, not any inverse of the skew.

Backwards — the rows of $C$ are the body-frame basis vectors expressed in inertial coordinates. The entry $C_{ij}={\hat{b}}_i\cdot {\hat{n}}_j$ makes this explicit.

Because $C$ is locked onto the curved surface $C{C}^T=I$; the only velocities that keep it on that surface point along the surface's tangent directions, which are precisely {skew matrix} $\times C$. Any other form would drag $C$ off the rotation group.

A rotation matrix, despite 9 entries, has only 3 true degrees of freedom (orthonormality removes 6). So 3 numbers in $\vec{\omega }$ exactly match the 3 ways $C$ can change — a perfect count from the Lie-algebra dimension, see Rotation group SO(3) and Lie algebra so(3).

The physical spin is fixed; the sign only records whether you wrote the map as $\mathcal{N}\to \mathcal{B}$ (minus) or $\mathcal{B}\to \mathcal{N}$ (plus). Both predict identical physical motion — you just must declare which convention you use.

Quaternions and Euler angles are just different coordinates for the same rotation; their rate equations (Quaternion kinematics — q̇ = ½ Ω(ω) q, Euler angle kinematics & gimbal lock) are re-parametrizations of this single DCM equation. Same physics, different bookkeeping.

Because Poisson's equation for rotating frames is the general statement that a body-fixed quantity's inertial rate equals $\vec{\omega }\times (\cdot )$; applied to each column/basis vector of the DCM it collapses to exactly $\dot{C}=-[\omega \times ]C$.

$\vec{\omega }$ is a rate, encoded in how $C$ is changing, not in its instantaneous value. You need $\dot{C}$ (or two nearby samples of $C$) to form $-\dot{C}{C}^T$ and read the spin off.

$\dot{C}=0$: the map is momentarily frozen. But $C$ itself can be any rotation — a non-spinning body can still be pointing in any fixed orientation.

It is a planar rotation about that fixed axis, given by the matrix exponential $e^{-[\omega \times ]t}C(0)$. The spin axis is an eigenvector with eigenvalue 1, so that axis's direction is preserved.

The simple $e^{-[\omega \times ]t}$ formula fails because skew matrices at different times don't commute; you must integrate the ODE step by step (or use a time-ordered product). Physically this is why real tumbling attitude has no closed form.

No — the DCM equation is perfectly well-behaved everywhere; only the Euler-angle re-parametrization becomes singular (rates blow up), see Euler angle kinematics & gimbal lock. This robustness is a key reason to propagate the DCM or a quaternion instead.

Nothing in the exact equation saturates — $C$ simply rotates faster, staying on SO(3). The practical limit is numerical: large $\omega$ times a finite step size causes big drift, forcing smaller steps or renormalization.

No — the equation only conserves $C{C}^T$; it does not restore it. If $C(0)C(0{)}^T\ne I$, that error is carried forward unchanged (its rate of change is zero, but so is its correction). Start clean.

No — the flow stays continuously on the connected component with $\det C=+1$. Since $\det$ is continuous and starts at $+1$, and the ODE never leaves $SO(3)$, it cannot jump to a reflection.