How the Lorentz transformation affects the metric tensor? After performing a Lorentz transformation, the orthogonal coordinates will become askew, as in the following figure:

and in such coordinate system, according to this Wikipedia article, the metric will have off-diagonal non-zero elements:
$$g_{ij}=\mathbf{e}_i.\mathbf{e}_j$$
which is not that of a flat spacetime:
$$ g = \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix}$$
What's the problem?
 A: You need to be very careful. The $\mathbf{e}_i$ are vectors, so they have a Lorentz index: $\mathbf{e}_i^\mu.$ When you write
$$\mathbf{e}_i \cdot \mathbf{e}_j$$
you actually mean
$$g_{\mu \nu} \mathbf{e}_i^\mu  \mathbf{e}_j^\nu$$
where $g_{\mu \nu}$ is the flat Minkowski metric (not the Euclidean metric). Once you know this, it's straightforward to check that the flat metric is preserved under Lorentz transformations.
Edit: in the theory of relativity, physicists refer to these objects as a frame or a vierbein. Might help you look for other resources if you want to.
A: Vibert is, of course, completely correct.  I'm gong to propose a slightly more geometric version of what he says.  The minkowski metric tensor is given by:
$$ds^{2} = g_{ab}dx^{a}dx^{b} = -dt^{2} + dx^{2} + dy^{2} + dz^{2}$$
Now, knowing that $v = \tanh\phi$, it is easy enough to show that the Lorentz transformations are given by:
$$\begin{align}
t' &= t \cosh \phi - x \sinh\phi\\
x' &= x \cosh \phi - t \sinh\phi
\end{align}$$
Taking the differential, solving for $dt$ and $dx$ and substituting into the line element above, we find:
$$ds'^{2} = -\left(\cosh^{2}\phi - \sinh^{2}\phi\right)dt'^{2} + \left(\cosh^{2}\phi - \sinh^{2}\phi\right)dx'^{2} + dy^{2} + dz^{2}$$
Since we know the basic trigonometric identity $\cosh^{2} - \sinh^{2} = 1$, we find that 
$$ds'^{2} = g_{ab}'dx'^{a}dx'^{b} = -dt'^{2} + dx'^{2} + dy^{2} + dz^{2}$$
and it is clear that the components of the metric tensor are the same.
EDIT: to see how this becomes the ordinary lorentz transform:
$$\begin{align}
\tanh\phi &= v\\
\frac{\sinh\phi}{\cosh\phi} & = v\\
\sinh^{2}\phi &= v^{2} \cosh^{2}\phi\\
\cosh^{2}\phi -1 &= v^{2}\cosh^{2}\phi\\
\cosh^{2}\phi\left(1 - v^{2}\right) &= 1\\
\cosh\phi &= \frac{1}{\sqrt{1-v^{2}}}\\
\sinh\phi &= \sqrt{cosh^{2}\phi -1}\\
&= \sqrt{\frac{1}{1-v^{2}} - 1}\\
&= \sqrt{\frac{1 - (1-v^{2}) }{1-v^{2}}}\\
&= \frac{v}{\sqrt{1-v^{2}}}
\end{align}$$
You can figure out the rest
A: The transformation law for a covariant vector $e$ (like a basis vector),   is :
$$e_i = \frac{\partial x^\mu}{\partial x^i} e_\mu \tag{1}$$
For a 2-covariant tensor like the metrics $g$, the transformation law is  :
$$g_{ij} = \frac{\partial x^\mu}{\partial x^i}  \frac{\partial x^\nu}{\partial x^j} g_{\mu\nu} \tag{2}$$
This simply means that $g_{ij}$ transforms as $e_i e_j$
The relation between $e_i^\mu$ and the partial derivatives is simply : 
$$e_i^\mu = \frac{\partial x^\mu}{\partial x^i}\tag{3}$$
So, you may write : 
$$e_i = e_i^\mu e_\mu, \qquad  g_{ij} =  e_i^\mu e_j^\nu g_{\mu\nu}\tag{4}$$
So, $e_i^\mu$ is the component of the new basis vector $e_i$, in the old basis vector $e_\mu$
A: The problem comes from the Euclidean representation of the orthogonality of two vectors in Minkowskian space: the scalar product of two vectors x,y with coordinates $x_{i},y_{i}$ (in 2D) is defined in the sense of Minkowski geometry by $xy=x_{1}y_{1}-x_{2}y_{2}$.
The two vectors are perpendicular to each other in the Minkowski sense if  $xy=x_{1}y_{1}-x_{2 }y_{2}=0$ , in the Euclidean sense, this equality translates the symmetry of the directions of the two vectors with respect to the bisectors of the coordinate angles. In particular, two vectors located on the same bisector are perpendicular to each other in Minkowski geometry.
A: I think the diagram may be confusing you. That diagram does not show anything "askew". The $x'$ and $ct'$ axes are orthogonal here in the spacetime sense. On the diagram (with scales chosen in the standard way) two 4-vectors are orthogonal if a line bisecting the angle between them goes at 45 degrees.
A: In one's rest frame, the x & ct axis are othogonal...x'-ct' are orthogonal to the observer who is moving with the primed frame (rest frame of primed observers)...but w.r.t. one frame the other frame do not have orthogonality...in fact invariance (see Proving that the Minkowski metric tensor is invariant under Lorentz transformations) of metric tensor diag. (1,-1,-1,-1) is justified from the former perspective...from the later perspective, it is like asking new components (primed) of metric tensor w.r.t. old basis & there u only get off diagonal term...all these are related to the illusion of drawing: trying to put 'time' coordinate in your 'paper' where subconsciously we assume spatial behaviour to time (in special relativity everything mixes but not on the same footing; see +; - both terms of metric): such idea clicked to me while seeing written notes beside a figure of Goldstein (cl. mech.; appropriate chapter)
