What is Minkowski spacetime? I was browsing through an article on spacetime when I caught the words Minkowski Spacetime.
A Wikipedia search brought me an article too complex for me to totally understand. So what is Minkowski spacetime?
 A: I am writing this answer to add some extra things to sahin's post, which he didn't expand upon.
By now you know that Minkowski space(time) is basically a four-dimensional vector space (space that is "flat", though obviously in lieu of metric structures, flatness is hard to define, though Minkowski space obviously does have a metric structure, the term "vector space" is well defined without a metric), however it is not the same as a four-dimensional euclidean vector space.
In 3-space, when you use orthogonal Descartes-coordinates, $x,y,z$, you can calculate the distance of a point $\mathbf{p}$ and $\mathbf{q}$, which have $(p_1,p_2,p_3)$ and $(q_1,q_2,q_3)$ coordinates by the pythagorean distance formula: $$ d(\mathbf{p,q})=\sqrt{(p_1-q_1)^2+(p_2-q_2)^2+(p_3-q_3)^2}=\sqrt{\Delta x_1^2+\Delta x_2^2+\Delta x_3^2}, $$ where the $\Delta x$s are the coordinate separations of points.
If you've learned some elementary vector geometry, basically what they teach in high schools is enough, then you will realize, this is basically the length of the vector $$ \mathbf{p}-\mathbf{q}, $$ which is calculated as $$\|\mathbf{p}-\mathbf{q}\|=\sqrt{(\mathbf{p}-\mathbf{q})\cdot(\mathbf{p}-\mathbf{q})}, $$ where the $\cdot$ denotes the scalar product.
If we perform a bit of an abstractation, we will see that the existence of a scalar product on a vector space, is what induces the familiar notions of geometry on that set. The definition of a lenth of a vector $\mathbf{x}$, is $$ \|\mathbf{x}\|=\sqrt{\mathbf{x}\cdot\mathbf{x}}, $$ and the angle between vectors $\mathbf{x}$ and $\mathbf{y}$ is $$ \alpha=\arccos\frac{\mathbf{x}\cdot\mathbf{y}}{\|\mathbf{x}\|\cdot\|\mathbf{y}\|}, $$ and these two terms are what makes concepts such as distances, angles and such make sense.
In short, and not very clearly, Minkowski space is a four-dimensional vector space, which has a scalar product, but this scalar product works a bit differently compared to the euclidean scalar product.
Let $\mathbb{R}^3$ denote the euclidean 3-space, and $\mathbb{M}$ the 4-dimensional Minkowski-space.
Let ${\mathbf{i},\mathbf{j},\mathbf{k}}\in\mathbb{R}^3$ be three, mutually orthogonal unit vectors pointing in the direction of coordinate axes $x,y,z$. In this case, you can expand all vectors in $\mathbb{R}^3$ with the help of these vectors, such as $$ \mathbf{u}=u_1 \mathbf{i}+u_2 \mathbf{j}+u_3 \mathbf{k} $$ and $$ \mathbf{v}=v_1 \mathbf{i}+v_2 \mathbf{j}+v_3 \mathbf{k}. $$ In this case, since the scalar product is distributive, you can write the scalar product of $\mathbf{u}$ and $\mathbf{v}$ as $$\mathbf{u}\cdot\mathbf{v}=u_1v_1(\mathbf{i}\cdot\mathbf{i})+u_1v_2(\mathbf{i}\cdot\mathbf{j})+u_1v_3(\mathbf{i}\cdot\mathbf{k})+u_2v_1(\mathbf{j}\cdot\mathbf{i})+u_2v_2(\mathbf{j}\cdot\mathbf{j})+u_2v_3(\mathbf{j}\cdot\mathbf{k})+u_3v_1(\mathbf{k}\cdot\mathbf{i})+u_3v_2(\mathbf{k}\cdot\mathbf{j})+u_3v_3(\mathbf{k}\cdot\mathbf{k}), $$ or in matrix form, as $$ \begin{pmatrix}
  u_1 & u_2 & u_3
 \end{pmatrix}
\begin{pmatrix}
  \mathbf{i}\cdot\mathbf{i} & \mathbf{i}\cdot\mathbf{j} & \mathbf{i}\cdot\mathbf{k} \\
  \mathbf{j}\cdot\mathbf{i} & \mathbf{j}\cdot\mathbf{j} & \mathbf{j}\cdot\mathbf{k} \\
  \mathbf{k}\cdot\mathbf{i} & \mathbf{k}\cdot\mathbf{j} & \mathbf{k}\cdot\mathbf{k}
 \end{pmatrix}
\begin{pmatrix}
  v_1 \\
  v_2 \\
  v_3 
 \end{pmatrix}. $$
Since the $\mathbf{i,j,k}$ vectors are mutually orthogonal, all scalar products of them where the two vectors do not coincide are 0, and the square matrix in the middle reduces to $$\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}, $$ which is the unit matrix, and thus to the usual formula for taking scalar products.
The reason I am doing the whole matrix shizzle is that I wish to emphasise, that you could, technically use DIFFERENT "basis vectors" compared to $\mathbf{i,j,k}$. You need them to subtend a nonzero volume, but they don't even have to be orthogonal or unit length. In that case, the square matrix in the middle would NOT reduce to a simple unit matrix, but some properties of the matrix would stay the same, for example the sign of its eigenvalues.
If you have not learned matrix theory, then the eigenvalues of a square matrix where only the diagonal has nonzero elements, are the diagonal elements themselves. For a square matrix, where there are nonzero elements in the off-diagonals, the eigenvalues are calculated in a bit more complicated way.
The euclidean scalar product satisfies certain properties, one of them is positive definiteness. That basically means, that if $\mathbf{x}\in\mathbb{R}^3$, then $$ \mathbf{x}\cdot\mathbf{x}\ge 0, $$ with equality if and only if $\mathbf{x}$ is the zero vector. The positive definitness of the scalar product implies that the matrix of the scalar product always has positive eigenvalues in ANY "basis".
In Minkowski space, the scalar product is NOT positive definite, but indefinite. The matrix of the scalar product has either one negative eigenvalue and three positive, or one positive and three negative. This is only a matter of convention.
You can also introduce a preferred basis in Minkowski space, four mutually orthogonal unit vectors, but one of them will have length in such a perverse way, that the square of its length is negative (assuming we follow the first convention. In this case, the matrix of the scalar product is $$\begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 &0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix} $$
Since, as I said above, all notion of geometry is induced by the scalar product, this means that the Minkowski space with the above scalar product is geometrically different than a 4-space with the usual euclidean scalar product.
What we call Minkowski space(time) then, is the pair $(\mathbb{R}^4,\eta)$, where $\eta$ is the indefinite Minkowski scalar product I described above.
A: 
As stated at http://en.wikipedia.org/wiki/Minkowski_space
"In 1905, with the publication in 1906, it was noted by Henri Poincaré that, by taking time to be the imaginary part of the fourth spacetime coordinate √−1 ct, a Lorentz transformation can be regarded as a rotation of coordinates in a four-dimensional Euclidean space with three real coordinates representing space, and one imaginary coordinate, representing time, as the fourth dimension.
Since the space is then a pseudo-Euclidean space, the rotation is a representation of a hyperbolic rotation, although Poincaré did not give this interpretation, his purpose being only to explain the Lorentz transformation in terms of the familiar Euclidean rotation. This idea was elaborated by Hermann Minkowski, who used it to restate the Maxwell equations in four dimensions, showing directly their invariance under the Lorentz transformation.
He further reformulated in four dimensions the then-recent theory of special relativity of Einstein. From this he concluded that time and space should be treated equally, and so arose his concept of events taking place in a unified four-dimensional space-time continuum.   In a further development, he gave an alternative formulation of this idea that did not use the imaginary time coordinate, but represented the four variables (x, y, z, t) of space and time in coordinate form in a four dimensional affine space."
However, its complete underlying meaning may still be a mystery for some folk.
It is far simpler for you to just figure out Special Relativity on your own. Intuitive common sense gets you to the destination. A simple mental analysis of "motion" leads you from ground zero to independently understanding Special Relativity, along with creating all of the important Special Relativity equations, meaning the Lorentz factor, Time dilation, Length contraction, Lorentz transformations, and Velocity addition, equations. As an outcome, you will in turn understand what Minkowski spacetime is all about.
Access to this analysis of "motion" can be found at my profile.
It is 1hr 39min long, which also means that I can obviously not text it here. This will provide a full description of relativity rather than be confined to posting of partial descriptions. If you can ride a bicycle, then you will understand the analysis.
A: I believe you know about the three dimensional Euclidean space which in general represented by $x, y$ and $z$ coordinates. Euclidean space has the important property of being flat. Now, you can think of the Minkowski space as a four dimensional space which is flat! The fourth coordinate is generally chosen to be time $t$, so people call it not "space" but "spacetime" in general. An important difference between Euclidean space and Minkowski spacetime occurs in the dependence on coordinates. That is; you have a coordinate dependence in Euclidean space whereas you are coordinate-free in Minkowski! So, you should realize that they have different space structures. Minkowski space is not an Euclidean space but it is instead called "pseudo-Euclidean" which is something different. I don't want to be too complicated in that, so I don't want to give mathematical definitions etc, but some reading on manifolds and tensors may help you.
And here is a good and simple article about what do we really mean by the word "flat". I think it will help you better than I do:
http://blogs.scientificamerican.com/degrees-of-freedom/2011/07/31/what-do-you-mean-the-universe-is-flat-part-ii/
A: 
What is Minkowski spacetime?

Being some particular sort of "spacetime", it is foremost some set of "events" (i.e. certain identifiable participants actually having met each other, or at least the thought-experimental idea of such a meeting of certain identifiable participants).
A set of events is a "Minkowski spacetime" due to satisfying further requirements, especially:


*

*of all participants who had met in any one event (actual, or thought up) each participated in other events ("before", or "after") as well,

*for each participant, the entire set events can be organized in three-dimensional "spatial hyperplanes (or subspaces)", which can likewise be ordered as "before", or "after", and

*all events were "flat" to each other, i.e. given the "interval" values $s^2$ between all pairs of events, which for a particular pair of events, $A$ and $B$, is
$~~~~~~~-$ either the duration, squared, of the trip of any participant travelling straight from $A$ to $B$ (or vice versa, from $A$ to $B$),
$~~~~~~~-$ or zero if the participants at event $B$ saw the meeting of participants at event $A$ (or vice versa, the participants at event $A$ saw event $B$),
$~~~~~~~-$  or otherwise the straight spatial distance,squared, between $A$ and $B$, divided by "$-c^2$",
then for any six events $A$, $B$, $J$, $K$, $P$, $Q$ the Cayley-Menger determinant
$$
\begin{vmatrix}
0 & s^2[~A, B~] & s^2[~A, J~] & s^2[~A, K~] & s^2[~A, P~] & s^2[~A, Q~] & 1~ \\
s^2[~A, B~] & 0 & s^2[~B, J~] & s^2[~B, K~] & s^2[~B, P~] & s^2[~B, Q~] & 1~ \\
s^2[~A, J~] & s^2[~B, J~] & 0 & s^2[~J, K~] & s^2[~J, P~] & s^2[~J, Q~] & 1~ \\
s^2[~A, K~] & s^2[~B, K~] & s^2[~J, K~] & 0 & s^2[~K, P~] & s^2[~K, Q~] & 1~ \\
s^2[~A, P~] & s^2[~B, P~] & s^2[~J, P~] & s^2[~K, P~] & 0 & s^2[~P, Q~] & 1~ \\ 
s^2[~A, Q~] & s^2[~B, Q~] & s^2[~J, Q~] & s^2[~K, Q~] & s^2[~P, Q~] & 0 & 1~ \\
1 & 1 & 1 & 1 & 1 & 1 & 0~
\end{vmatrix}$$
vanishes.
(Some, apparently including Minkowski himself, prefer to also sprinkle real number tuples as "coordinates" on these events.) 
