Show that if $\nabla { {\bf{u}}}+(\nabla { {\bf{u}})^T}=0$ then $\bf{u}(x)=\bf{a}\times \bf{x}+\bf{b}$ I am working on this problem:

Prove that if a smooth($C^1$) function $\bf{u}$ satisfies $\nabla { {\bf{u}}}+(\nabla { {\bf{u}})^T}=0$ then there exist constant vectors ${\bf{a},\bf{b}}\subset \Bbb R^3$, such that $\bf{u}(x)=\bf{a}\times \bf{x}+\bf{b}$, where $\bf{x} =<x_1,x_2,x_3>$.

I really have no clue about this problem and I am not familiar with matrix calculations. So I just tried to verify that $\bf{u}(x)=\bf{a}\times \bf{x}+\bf{b}$ actually satisfy $\nabla { {\bf{u}}}+(\nabla { {\bf{u}})^T}=0$ .
I tried to calculate $\nabla ({ {\bf{\bf{a}\times \bf{x}+\bf{b}}}}）$ and get $\bf{a}\times \nabla {\bf{x}}$. 
Then if I calculate $(\nabla ({ {\bf{\bf{a}\times \bf{x}+\bf{b}}}})^T$, is the result $(\bf{a}\times \nabla {\bf{x}})^T$? 
It doesn't seem right to me, since I cannot conclude $\bf{a}\times \nabla {\bf{x}}+(\bf{a}\times \nabla {\bf{x}})^T=0$.
Could anyone kindly provide some help? For example, some reference or theorem can be used here? Thanks very much！
 A: You are confusing divergence ($\nabla \cdot$) with gradient ($\nabla$).

The gradient can be used on vector functions as well, but this gives you are matrix-valued instead of an scalar-valued function.

Try to compute the first line of
$${ {\bf{\bf{a}\times \bf{x}+\bf{b}}}}.$$
which is
$$a_2x_3-a_3x_2+b_1$$
Then use the gradient on it. This gives you
$$\nabla (a_2x_3-a_3x_2+b_1) = \begin{pmatrix}0\\-a_3 \\a_2\end{pmatrix}$$
This gives you the first column (or row, not 100% certain, see bottom) of your $ \nabla { {\bf{\bf{a}\times \bf{x}+\bf{b}}}}$.
Do so for the rest.
$$\nabla ( {\bf{\bf{a}\times \bf{x}+\bf{b}}}) =\begin{pmatrix}0 &a_3&-a_2\\-a_3 & 0 & a_1\\a_2 &-a_1&0\end{pmatrix} (=:A)$$
The matrix you get, should be antisymmetric ($A^T = -A$), which it is.
This is exactly what you need to show.
IMPORTANT:
I'm not certain if $\nabla (a_2x_3-a_3x_2+b_1)$ should be $\begin{pmatrix}0\\-a_3 \\a_2\end{pmatrix}$ or $\begin{pmatrix}0 &-a_3 &a_2\end{pmatrix}$.
Depending on the convention you need to transpose everthing above.
A: Hint:
Once you figure out how to take the derivative of $a \times x$ properly, you'll have no trouble showing that if $u(x) = a \times x + b$, then it will satisfy $\nabla u + \nabla u^T = 0$.
Next: suppose that 
$$
u = (u_1(x_1,x_2,x_3),u_2(x_1,x_2,x_3),u_3(x_1,x_2,x_3))
$$
then $\nabla u + \nabla u^T = 0$ gives us the equations
$$
\frac{\partial u_i}{\partial x_j} + 
\frac{\partial u_j}{\partial x_i} = 0 \qquad i,j = 1,2,3
$$
