Why is the derivative of the quaternion equal to the Kronecker product of the quaternion itself and the angular velocity?

the dimension of $\dot{\mathrm{Q}}(t)$ is 4x1 but the following product is 4x4 $$\dot{\mathrm{Q}}(t)=\frac{1}{2}\mathrm{Q}(t)\otimes\overline{\Omega}(t)\ , (1)$$ equation (1) is part of the dynamic attitude system of rigid body

syms q0 q1 q2 q3 w1 w2 w3 real
Q=[q0 q1 q2 q3]' % quaternion
W=[w1 w2 w3]     % angular velocity
W_bar=[0 W]
Q_dot=0.5*kron(Q,W_bar)  % derivative of the quaternion

the result is :

Q =

q0
q1
q2
q3

W =

[ w1, w2, w3]

W_bar =

[ 0, w1, w2, w3]

Q_dot =

[ 0, (q0*w1)/2, (q0*w2)/2, (q0*w3)/2]
[ 0, (q1*w1)/2, (q1*w2)/2, (q1*w3)/2]
[ 0, (q2*w1)/2, (q2*w2)/2, (q2*w3)/2]
[ 0, (q3*w1)/2, (q3*w2)/2, (q3*w3)/2]
• @Qmechanic would you consider removing the complex-numbers tag? I've edited my answer to explain why these are not complex numbers. – user197851 Jul 24 '18 at 17:03
• @LonelyProf: From the tag wiki: "The complex-number tag includes quaternion, octonions,... " – Qmechanic Jul 24 '18 at 17:49

Your expression for quaternion multiplication is wrong. It is not the Kronecker (or outer) product, although some people use the same symbol $\otimes$ to represent it. Multiplying two 4-element quaternions together yields another 4-element quaternion. We need to take care with the sign convention, and the convention of body-fixed or space-fixed angular velocity, but I believe the formula you are looking for is (in matrix multiplication form) $$\begin{pmatrix} \dot{Q}_0 \\ \dot{Q}_1 \\ \dot{Q}_2 \\ \dot{Q}_3 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} Q_0 & -Q_1 & -Q_2 & -Q_3 \\ Q_1 & Q_0 & -Q_3 & Q_2 \\ Q_2 & Q_3 & Q_0 & -Q_1 \\ Q_3 & -Q_2 & Q_1 & Q_0 \end{pmatrix} \begin{pmatrix} 0 \\ \bar{\Omega}_1 \\ \bar{\Omega}_2 \\ \bar{\Omega}_3 \end{pmatrix}$$
Generally, if we write a quaternion as a combination of a scalar and a vector $\mathbf{A}= \bigl(a_0,\mathbf{a}\bigr)=\bigl(a_0,a_1,a_2,a_3\bigr)$, and similarly for $\mathbf{B}$ and $\mathbf{C}$, then the quaternion product may be expressed $$\mathbf{C} = (c_0, \mathbf{c}) = \mathbf{A}\otimes\mathbf{B} = \bigl( a_0b_0-\mathbf{a}\cdot\mathbf{b},a_0\mathbf{b}+b_0\mathbf{a}+\mathbf{a}\times\mathbf{b} \bigr)$$ where $\cdot$ is the usual scalar product of two vectors, and $\times$ is the vector cross product. If we expand this into individual terms, we can see that it may be written in a similar form to my answer above $$\begin{pmatrix} c_0 \\ c_1 \\ c_2 \\ c_3 \end{pmatrix} = \begin{pmatrix} a_0 & -a_1 & -a_2 & -a_3 \\ a_1 & a_0 & -a_3 & a_2 \\ a_2 & a_3 & a_0 & -a_1 \\ a_3 & -a_2 & a_1 & a_0 \end{pmatrix} \begin{pmatrix} b_0 \\ b_1 \\ b_2 \\ b_3 \end{pmatrix}$$