How to invert $ E=B \wedge v$? I'm considering electromagnetic wave. If I have $E(x,y,z)$ how to get $B(x,y,z)$ . 
I think it's such a chilly question but I don't know how to invert
$$ E=B \wedge v$$
It's a cross product so I do not think it's trivial. It's anti-commutative so the order is important.
thank you so much
 A: The relation you're trying to invert likely comes from the Ampère-Maxwell law, which in vacuum reads
$$
\nabla \times\mathbf B = \frac{1}{c^2}\frac{\partial \mathbf E}{\partial t}
$$
in SI units. For plane waves, if you write $\mathbf E = \mathrm{Re}(\mathbf E_0 e^{i(\mathbf k\cdot \mathbf r-\omega t)})$ and $\mathbf B = \mathrm{Re}(\mathbf B_0 e^{i(\mathbf k\cdot \mathbf r-\omega t)})$, the Ampère-Maxwell law reduces to 
$$
i\mathbf k\times\mathbf B_0 = \frac{-i\omega}{c^2}\mathbf E_0,
$$
and if you then set $\mathbf v = \frac{c^2}{\omega}\mathbf k$, i.e. the vector along $\mathbf k$ with magnitude $c$, then that reads
$$
\mathbf E_0 = \mathbf B_0 \times\mathbf v.
$$

That relationship gives you the electric field amplitude $\mathbf E_0$ if you know the magnetic field amplitude $\mathbf B_0$, but it can't (easily*) be inverted. If you want to go the other way, the thing to do is to use the other curl equation in the Maxwell set, i.e. the Faraday law
$$
\nabla \times\mathbf E = -\frac{\partial \mathbf B}{\partial t}
$$
which for plane waves reads
$$
i\mathbf k\times\mathbf E_0 = i\omega\mathbf B_0,
$$
and with the same definition of $\mathbf v = \frac{c^2}{\omega}\mathbf k$, gives you
$$
\mathbf B_0 = \frac{1}{c^2}\mathbf v \times\mathbf E_0,
$$
which is presumably the relation you were looking for. However, it's important to keep in mind that this is only true in SI units, and if you're working in a separate unit set then you need to repeat the derivation starting from the correct version of the Maxwell equations.
Similarly, since your question is severely lacking in context, this answer assumes that you're working in vacuum. If you're working in a linear medium, substitute in the refractive index where appropriate.

* If you really insist on directly inverting the relationship $\mathbf E_0 = \mathbf B_0 \times\mathbf v$ then it's possible, because Gauss's law tells you that $\mathbf v\cdot \mathbf B_0=0$, which means that you can take the cross product of your relation with $\mathbf v$ to give you 
\begin{align}
\mathbf v\times\mathbf E_0 
& = \mathbf v\times(\mathbf B_0\times\mathbf v)
\\ & = \mathbf B_0 (\mathbf v \cdot\mathbf v) - \mathbf v (\mathbf v \cdot\mathbf B_0) 
\\ & = c^2 \mathbf B_0,
\end{align} matching the previous handling. However, the proper way to go about this, really, is to just go back up to the Faraday law.
A: I am not 100% sure what you are trying to do, but I can still answer purely mathematically how one would handle this in 3D space. Let me assume that one somehow gets the expression $E = B \times v$ for some vectors $E, B, v$ which may not have any relationship to the normal electric and magnetic fields; and let me assume that $\times$ is the normal cross product in 3D.
First, define $c = B \cdot v$ for the normal dot product in 3D, then use it to define the perpendicular component $\bar B = B - c~v.$ Note that $E = \bar B \times v$ as well, but now $v \cdot \bar B = 0.$
One may then use the BAC-CAB rule,
$$ A\times (B \times C) = B(A\cdot C) - C(A \cdot B),$$
which means that $v \times E = v \times (\bar B \times v) = \bar B (v \cdot v) - v (v \cdot \bar B).$ The last term of course was carefully set to 0.
This means that one now has $$B = c~v + \frac1{v^2} v\times E.$$
If you only know $E$ but not $c$ then you can view this as saying "there exists a family of possible $B$ parameterized by $c$ which all solve this equation."
(If you are interested in how this generalizes to larger-dimensional spaces and orientations, the wedge product in component form becomes this big antisymmetric tensor $\epsilon^{\alpha\beta\dots\omega}$ and we are basically looking in the 3D case of $\epsilon^{\alpha\lambda\mu} \epsilon_{\alpha\nu\omicron}$ that we must have two full permutations of all the indices, so we must either have $\lambda = \nu$ and $\mu = \omicron$ or $\lambda = \omicron$ and $\mu = \nu$. In one of these the two permutations have the same parity giving a +1; in the other the two permuations have opposite parity giving a -1; so one gets $\delta^\lambda_\nu~\delta^\mu_\omicron - \delta^\lambda_\omicron~\delta^\mu_\nu.$ So in, say, 4D you presumably have three-$\delta$ terms, presumably 9 in all.)
A: If you have an electromagnetic wave, with a corresponding electric field $\vec E (x,y,z,t)$, one can find the magnetic field through Faraday's law, namely,
$$\nabla \times \vec E = -\frac{\partial \vec B}{\partial t}$$
supplemented by an initial condition for $\vec B$ as it is a system of first-order differential equations. From Maxwell's equations, one may derive the electromagnetic wave equations, and so the $\vec B$ field obtained by Faraday's law will also satisfy a wave equation,
$$\nabla^2 \vec B = \mu_0 \epsilon_0 \frac{\partial^2 \vec B}{\partial t^2}.$$

Example
Suppose we have an electric field, $\vec E = E_0 \sin(ky-\omega t) \hat x$. The curl is given by,
$$\nabla \times \vec E = -E_0k \cos(ky-\omega t)\hat z.$$
Integrating with respect to $t$, one finds,
$$\vec B =E_0 k \int \mathrm dt \, \cos(ky-\omega t) \hat z = -\frac{E_0 k}{\omega}\sin(ky-\omega t) \hat z + C.$$
If we demand say, at time $t=0$ and at $y=0$, $\vec B = 0$ then $C= 0$. This of course also satisfies the wave equation, as
$$\frac{\partial^2 \vec B}{\partial y^2} = -k^2 \sin (ky)\hat z = \frac{\partial^2 \vec B}{\partial t^2} = \mu_0 \epsilon_0 \omega^2 \sin(ky-\omega t) \hat z$$
providing that $c^{-2} = \mu_0 \epsilon_0$.
A: If $\vec E=\vec B\times\vec v$, then the electric field components are $E_i=\epsilon_{ijk}B_jv_k=M_{ij}B_j$, where $M_{ij}=\epsilon_{ijk}v_k$. In matrix notation
$$\vec E=M\vec B.$$
Let us assume there is an inverse $M^{-1}$. Multiply it on the left and obtain
$$\vec B=M^{-1}\vec E.$$
The problem has been reduced to finding an inverse of a $3\times 3$ matrix. However, since 
$$M=\begin{bmatrix}0&v_3&-v_2\\-v_3&0&v_1\\v_2&-v_1&0\end{bmatrix},$$
and $\det M=0$, the matrix is singular and the inverse $M^{-1}$ does not exist! 
