PART A : Plane Phase Waves and Lorentz Transformations
In the following it's necessary to use a more general form of Lorentz Transformation between two systems $\:S\:$ and $\:S^{\prime}\:$. So, let the system $\:S\:$ translating with constant velocity $\:\mathbf{v}\:$ with respect to $\:S^{\prime}\:$ (see Figure in the bottom). With finite variables the equations of this general Lorentz Transforma- tion are
\begin{align}
\mathbf{x} & = \mathbf{x}^{\boldsymbol{\prime}}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \mathbf{x}^{\boldsymbol{\prime}})\mathbf{n}-\gamma \mathbf{v}t^{\boldsymbol{\prime}}
\tag{A-01a}\\
t & = \gamma\left(t^{\boldsymbol{\prime}}-\dfrac{\mathbf{v}\boldsymbol{\cdot} \mathbf{x}^{\boldsymbol{\prime}}}{c^{2}}\right)
\tag{A-01b}
\end{align}
In differential form
\begin{align}
\mathrm{d} \mathbf{x} & = \mathrm{d}\mathbf{x}^{\boldsymbol{\prime}}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \mathrm{d} \mathbf{x}^{\boldsymbol{\prime}})\mathbf{n}-\gamma \mathbf{v} \mathrm{d} t^{\boldsymbol{\prime}}
\tag{A-02a}\\
\mathrm{d} t & = \gamma\left( \mathrm{d} t^{\boldsymbol{\prime}}-\dfrac{\mathbf{v}\boldsymbol{\cdot} \mathrm{d} \mathbf{x}^{\boldsymbol{\prime}}}{c^{2}}\right)
\tag{A-02b}
\end{align}
Dividing equations (A-02) side by side the velocity 3-vector of a particle is transformed as
\begin{equation}
\mathbf{u} = \dfrac{\mathbf{u}^{\boldsymbol{\prime}}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \mathbf{u}^{\boldsymbol{\prime}})\mathbf{n}-\gamma \mathbf{v}}{\gamma \Biggl(1-\dfrac{\mathbf{v}\boldsymbol{\cdot} \mathbf{u}^{\boldsymbol{\prime}}}{c^{2}}\Biggr)}
\tag{A-03}
\end{equation}
Equations for the inverse transforms are produced by transposition of unprimed with primed variables and $\:\mathbf{v}\:$ with $\:-\mathbf{v}\:$
\begin{align}
\mathbf{x}^{\boldsymbol{\prime}} & = \mathbf{x}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \mathbf{x})\mathbf{n}+\gamma \mathbf{v} t
\tag{A-04a}\\
t^{\boldsymbol{\prime}}& = \gamma\left( t+\dfrac{\mathbf{v}\boldsymbol{\cdot} \mathbf{x}}{c^{2}}\right)
\tag{A-04b}
\end{align}
\begin{align}
\mathrm{d} \mathbf{x}^{\boldsymbol{\prime}} & = \mathrm{d}\mathbf{x}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \mathrm{d} \mathbf{x})\mathbf{n}+\gamma \mathbf{v} \mathrm{d} t
\tag{A-05a}\\
\mathrm{d} t ^{\boldsymbol{\prime}}& = \gamma\left( \mathrm{d} t+\dfrac{\mathbf{v}\boldsymbol{\cdot} \mathrm{d} \mathbf{x}}{c^{2}}\right)
\tag{A-05b}
\end{align}
\begin{equation}
\mathbf{u}^{\boldsymbol{\prime}} = \dfrac{\mathbf{u}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \mathbf{u})\mathbf{n}+\gamma \mathbf{v}}{\gamma \Biggl(1+\dfrac{\mathbf{v}\boldsymbol{\cdot} \mathbf{u}}{c^{2}}\Biggr)}
\tag{A-06}
\end{equation}
Now let a plane wave characterized by a scalar invariant quantity $\:\phi\:$, the phase
\begin{equation}
\phi \left( \mathbf{x},t\right)= \omega \; t -\mathbf{k}\boldsymbol{\cdot}\mathbf{x}
\tag{A-07}
\end{equation}
In (A-07)
\begin{equation}
\omega= 2\pi\nu
\tag{A-08}
\end{equation}
is the angular frequency and $\;\nu\;$ the frequency. Also
\begin{equation}
\mathbf{k}= \dfrac{ 2\pi}{\lambda} \;\mathbf{m} , \qquad \mathbf{m} =\dfrac{ \mathbf{k}}{\Vert \mathbf{k}\Vert }
\tag{A-09}
\end{equation}
is the wave 3-vector and $\;\lambda\;$ the wavelength.
The plane wave "propagates" with velocity vector
\begin{equation}
\mathbf{w}= \dfrac{ \omega}{\Vert \mathbf{k}\Vert } \;\mathbf{m}=\lambda\nu\;\mathbf{m} = \dfrac{ \omega}{\Vert \mathbf{k}\Vert^{2}}\mathbf{k}, \qquad \Vert \mathbf{w}\Vert \equiv \mathrm{w} = \dfrac{ \omega}{\Vert \mathbf{k}\Vert }=\lambda\nu
\tag{A-10}
\end{equation}
To see how this phase is transformed in the primed system $\:S^{\prime}\:$, we insert the expressions (A-01) in (A-07)
\begin{align}
\phi \left( \mathbf{x},t\right) & = \omega \; t -\mathbf{k}\boldsymbol{\cdot}\mathbf{x}\\
\phi \left( \mathbf{x},t\right) & = \omega \underbrace{\left[\gamma\left(t^{\boldsymbol{\prime}}-\dfrac{\mathbf{v}\boldsymbol{\cdot} \mathbf{x}^{\boldsymbol{\prime}}}{c^{2}}\right)\right]}_{t} -\mathbf{k}\boldsymbol{\cdot} \underbrace{\biggl[\mathbf{x}^{\boldsymbol{\prime}}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \mathbf{x}^{\boldsymbol{\prime}})\mathbf{n}-\gamma \mathbf{v}t^{\boldsymbol{\prime}}\biggr]}_{\mathbf{x}}\\
& = \underbrace{\biggl[ \gamma\left(\omega+\mathbf{k}\boldsymbol{\cdot}\mathbf{v}\right)\biggr]}_{\omega ^{\boldsymbol{\prime}}} t^{\boldsymbol{\prime}} - \underbrace{\biggl[\mathbf{k}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \mathbf{k})\mathbf{n}+\gamma \mathbf{v} \dfrac{\omega}{c^{2}}\biggr]}_{\mathbf{k}^{\boldsymbol{\prime}}}\boldsymbol{\cdot} \mathbf{x}^{\boldsymbol{\prime}}\\
& = \omega^{\boldsymbol{\prime}} t^{\boldsymbol{\prime}} - \mathbf{k}^{\boldsymbol{\prime}}\boldsymbol{\cdot} \mathbf{x}^{\boldsymbol{\prime}}=\phi ^{\boldsymbol{\prime}}\left( \mathbf{x}^{\boldsymbol{\prime}},t^{\boldsymbol{\prime}} \right)
\tag{A-11}
\end{align}
so
\begin{equation}
\phi ^{\boldsymbol{\prime}}\left( \mathbf{x}^{\boldsymbol{\prime}},t^{\boldsymbol{\prime}} \right)= \omega^{\boldsymbol{\prime}} t^{\boldsymbol{\prime}} - \mathbf{k}^{\boldsymbol{\prime}}\boldsymbol{\cdot} \mathbf{x}^{\boldsymbol{\prime}}
\tag{A-12}
\end{equation}
where
\begin{align}
\mathbf{k}^{\boldsymbol{\prime}} & = \mathbf{k}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \mathbf{k})\mathbf{n}+\gamma \mathbf{v} \dfrac{\omega}{c^{2}}
\tag{A-13a}\\
\omega^{\boldsymbol{\prime}} & = \gamma\left(\omega+\mathbf{v}\boldsymbol{\cdot}\mathbf{k}\right)
\tag{A-13b}
\end{align}
Multiplying (A-13a) and (A-04b) with c, equations (A-13) and (A-04) are rewritten
\begin{align}
c\mathbf{k}^{\boldsymbol{\prime}} & = c\mathbf{k}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} c\mathbf{k})\mathbf{n}+\gamma \mathbf{v} \dfrac{\omega}{c}
\tag{A-14a}\\
\omega^{\boldsymbol{\prime}} & = \gamma\left(\omega+ \dfrac{ \mathbf{v}\boldsymbol{\cdot}c\mathbf{k}}{c}\right)
\tag{A-14b}
\end{align}
\begin{align}
\mathbf{x}^{\boldsymbol{\prime}} & = \mathbf{x}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \mathbf{x})\mathbf{n}+\gamma \mathbf{v} \dfrac{ct}{c}
\tag{A-15a}\\
c t^{\boldsymbol{\prime}}& = \gamma\left( ct+\dfrac{\mathbf{v}\boldsymbol{\cdot} \mathbf{x}}{c}\right)
\tag{A-15b}
\end{align}
that is the vector
\begin{equation}
\boldsymbol{\Omega} \equiv \left(\omega,c\mathbf{k} \right)
\tag{A-16}
\end{equation}
is Lorentz transformed as the space-time position 4-vector
\begin{equation}
\mathbf{X} = \left(ct,\mathbf{x} \right)
\tag{A-17}
\end{equation}
The phase is the invariant inner product "$\:\bullet\:$" of two 4-vectors in Minkowski space
\begin{equation}
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\phi ^{\boldsymbol{\prime}}\left( \mathbf{x}^{\boldsymbol{\prime}},t^{\boldsymbol{\prime}}\right)= \omega^{\boldsymbol{\prime}} t^{\boldsymbol{\prime}} - \mathbf{k}^{\boldsymbol{\prime}}\boldsymbol{\cdot} \mathbf{x}^{\boldsymbol{\prime}}
= \dfrac{1}{c} \left(\boldsymbol{\Omega}^{\boldsymbol{\prime}}\bullet\mathbf{X}^{\boldsymbol{\prime}} \right)= \dfrac{1}{c} \left(\boldsymbol{\Omega}\bullet\mathbf{X} \right)= \omega \; t -\mathbf{k}\boldsymbol{\cdot}\mathbf{x}=\phi \left( \mathbf{x},t\right)
\tag{A-18}
\end{equation}
PART B : A Lorentz invariant picture of a "subluminal" particle and its accompaning "superluminal" plane phase wave.
Multiplying (A-14a) by c and dividing side by side with (A-14b) we have, under the assumption $\:\omega \ne 0 \ne \omega^{\boldsymbol{\prime}}$, that
\begin{equation}
\left(c^{2} \mathbf{k}^{\boldsymbol{\prime}}/\omega^{\boldsymbol{\prime}}\right)= \dfrac{\left(c^{2} \mathbf{k}/\omega\right) +(\gamma-1)\biggl[\mathbf{n}\boldsymbol{\cdot}\left(c^{2} \mathbf{k}/\omega\right) \biggr] \mathbf{n}+\gamma \mathbf{v}}{\gamma \Biggl[1+\dfrac{\mathbf{v}\boldsymbol{\cdot}\left(c^{2} \mathbf{k}/\omega\right) }{c^{2}}\Biggr]}
\tag{B-01}
\end{equation}
Above equation (B-01) is identical to (A-06) which is repeated here for comparison
\begin{equation}
\mathbf{u}^{\boldsymbol{\prime}} = \dfrac{\mathbf{u}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \mathbf{u})\mathbf{n}+\gamma \mathbf{v}}{\gamma \Biggl(1+\dfrac{\mathbf{v}\boldsymbol{\cdot} \mathbf{u}}{c^{2}}\Biggr)}
\tag{A-06}
\end{equation}
So, for a plane phase wave with angular frequency 4-vector $\;\boldsymbol{\Omega} \equiv \left(\omega,c\mathbf{k} \right) \;$
the velocity 3-vector $\;\mathbf{u}=c^{2} \mathbf{k}/\omega\;$ is transformed as the velocity 3-vector $\;\mathbf{u}\;$ of a particle.
But in order for this velocity to correspond to a "subluminal" particle we must have
\begin{equation}
\mathrm{u}\equiv \Vert \mathbf{u} \Vert =\left\Vert \dfrac{c^{2} \mathbf{k}}{\omega}\right\Vert = \dfrac{c^{2}}{\mathrm{w}} < c
\tag{B-02}
\end{equation}
From above equation we conclude that :
- The plane phase wave must be "superluminal" : $\;\mathrm{w}>c$. This is identical to time-like $\;\boldsymbol{\Omega} \;$:
\begin{equation}
\Vert \boldsymbol{\Omega} \Vert^{2} = \omega ^{2} - c^{2} \Vert \mathbf{k} \Vert^{2} >0
\tag{B-03}
\end{equation}
- The picture of this "superluminal" plane phase wave and its accompanied "subluminal" particle is Lorentz invariant. Note that a "superluminal" plane phase wave keeps this property in all frames since $\:\Vert \boldsymbol{\Omega} \Vert^{2}\left(>0\right)\:$ is invariant.
- The product of the speed of the particle by the speed of the wave is invariant
\begin{equation}
\mathrm{u}^{\boldsymbol{\prime}}\cdot\mathrm{w}^{\boldsymbol{\prime}}=c^{2}=\mathrm{u}\cdot\mathrm{w}
\tag{B-04}
\end{equation}
PART C : Louis de Broglie hypothesis and waves
In PART B we associate to a "superluminal" plane phase wave a "subluminal" particle. In the analysis therein we produce from two characteristics of the wave, that is the frequency $\;\omega\;$ and the wave vector $\;\mathbf{k}\;$, the velocity of the particle. But by the velocity only the particle is not defined completely. We must define,for example, its rest mass or its energy content at least in one frame. And reversely, by the velocity of a particle only we can find the velocity of the wave but not its frequency or its wavelength. Something is missing. But since in the first case we can't find the energy content of the particle and in the second case the frequency of the wave, the "missing something" must be a relation between these two quantities.
Le's go to the rest frame of the particle. Any quantity with respect to this frame will be subscripted with an $\:_{\mathrm{o}}\;$. In this frame $\;\mathrm{u}_{\mathrm{o}}=0\;$, so if we want to keep the invariance (B-04) in this limit we must have $\;\mathrm{w}_{\mathrm{o}}=\infty \;$. The wave would have wavelength $\;\lambda_{\mathrm{o}}=\infty\;$, so it would represent a uniform in space periodic in time vibration with frequency let $\;\nu_{\mathrm{o}}$. Note than this is in agreement with a "superluminal" plane phase wave, that is a time-like $\;\boldsymbol{\Omega} \equiv \left(\omega,c\mathbf{k} \right) \;$. We can find an inertial frame in which the space component of a time-like 4-vector would be zero
\begin{equation}
\boldsymbol{\Omega}_{\mathrm{o}} = \left(\omega_{\mathrm{o}},\mathbf{0} \right) =\left(2\pi\nu_{\mathrm{o}},\mathbf{0} \right)
\tag{C-01}
\end{equation}
At this point exactly we meet the de Broglie hypothesis : Since to a photon with energy $\;E\;$ there corresponds a frequency
\begin{equation}
\nu = \dfrac{E}{h}
\tag{C-02}
\end{equation}
then to a massive particle with rest mass $\;m_{\mathrm{o}}\;$ or energy in its rest frame $\;E_{\mathrm{o}} =m_{\mathrm{o}}c^{2}\;$
there corresponds a wave in its rest frame with frequency
\begin{equation}
\nu_{\mathrm{o}} = \dfrac{E_{\mathrm{o}} }{h}= \dfrac{m_{\mathrm{o}}c^{2}}{h}
\tag{C-03}
\end{equation}
So in a sense the particle is like to have an internal oscillation of frequency $\;\nu_{\mathrm{o}}$.
In the rest frame of the particle the momentum-energy 4-vector is
\begin{equation}
c\mathbf{P}_{\mathrm{o}}= \left(E_{\mathrm{o}},\mathbf{0} \right)
\tag{C-04}
\end{equation}
which combined with (C-01) and (C-03) gives
\begin{equation}
c\mathbf{P}_{\mathrm{o}}-\hbar \boldsymbol{\Omega}_{\mathrm{o}} = \left(0,\mathbf{0} \right)
\tag{C-05}
\end{equation}
or more clearly
\begin{equation}
\left(c\mathbf{P}-\hbar \boldsymbol{\Omega}\right)_{\mathrm{o}} = \left(0,\mathbf{0} \right)
\tag{C-06}
\end{equation}
that is there exists a frame (the rest of the particle) where the 4-vector $\; \left(c\mathbf{P}-\hbar \boldsymbol{\Omega}\right)\;$ is the zero 4-vector. So, if we apply a Lorentz Transformation to both sides of (C-06) then in any inertial frame
\begin{equation}
c\mathbf{P}-\hbar \boldsymbol{\Omega} = \left(0,\mathbf{0} \right)
\tag{C-07}
\end{equation}
since the transform of the zero 4-vector $\;\left(0,\mathbf{0} \right) \;$ is this itself. So in any frame
\begin{equation}
c\mathbf{P}=\hbar \boldsymbol{\Omega}
\tag{C-08}
\end{equation}
written explicitly with the time and space components
\begin{equation}
\left(E,c\mathbf{p}\right)=\left(\hbar \omega,\hbar c \mathbf{k} \right)
\tag{C-09}
\end{equation}
Equating time and space components
\begin{align}
E & =\hbar \omega=h\nu
\tag{C-10a}\\
\mathbf{p} & = \hbar \mathbf{k}
\tag{C-10b}
\end{align}
Equation (C-10a) is expressed as
\begin{equation}
\nu = \dfrac{E }{h}= \dfrac{\sqrt{\left(m_{\mathrm{o}}c^{2}\right)^{2}+\left(pc\right)^{2}}}{h}=\dfrac{\gamma m_{\mathrm{o}}c^{2}}{h}
\tag{C-11}
\end{equation}
and (C-10b) as
\begin{equation}
\lambda = \dfrac{h }{p}=\dfrac{h}{\gamma m_{\mathrm{o}}v}
\tag{C-12}
\end{equation}
so
\begin{equation}
\mathrm{w}=\lambda \,\nu=\dfrac{c^{2}}{v}
\tag{C-13}
\end{equation}
See a larger size animation of this particle and its wave here : http://i.imgur.com/hsL0OfM.gifv
Note :
The general Lorentz Transformation (A-01) is given without proof in "CLASSICAL ELECTRODYNAMICS" by J.D.Jackson, 3rd Edition , $\S$ 11.3, but with the unprimed and primed systems interchanged as follows
\begin{align}
x'_{0} & =\gamma\left(x_{0}-\boldsymbol{\beta}\boldsymbol{\cdot}\mathbf{x}\right)\\
\mathbf{x}^{\prime} & = \mathbf{x} +\dfrac{\left(\gamma-1\right)}{\beta^{2}}\left(\boldsymbol{\beta}\boldsymbol{\cdot}\mathbf{x} \right)\boldsymbol{\beta}-\gamma\boldsymbol{\beta}x_{0}
\tag{11.19}
\end{align}
\begin{align}
\boldsymbol{\beta} & = \dfrac{\mathbf{v}}{c}\; \qquad \beta=|\boldsymbol{\beta}| \\
\gamma &=\left(1-\beta^2 \right)^{-1/2}
\tag{11.17}
\end{align}
A derivation of this transformation is given in my answer as user82794 therein : Two sets of coordinates each in frames $O$ and $ O' $ (Lorentz transformation), SECTION B.
