Take $c=1$. I will be using the metric signature of $(-,+,+,+)$, since I am learning from Helliwell and Sahakian. Kevin Zhou (and many other textbooks) uses $(+,-,-,-)$ though.
The coordinates of a frame $(t',x')$ moving in the $+x$ direction at speed $v$ is related to the coordinates of the stationary frame $(t,x)$ by the Lorentz transform:
$$ t'=\gamma(t-vx) \\ x'=\gamma(x-vt) $$
assuming at $t=t'=0$ the origins of the frames coincide. Alternatively you can think of this as the temporal and spatial separation between two events (the delta in front of $t$ and $x$ is then implied)
We can write this in 3 dimensions as a rotation matrix:
$$ \begin{pmatrix} t \\ x \\ y \\ z \end{pmatrix} = \begin{pmatrix} \gamma & \gamma v & 0 & 0 \\ \gamma v & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} t' \\ x' \\ y' \\ z' \end{pmatrix} \implies \mathbf r = \hat {\boldsymbol \Lambda} \cdot \mathbf r' $$
Einstein summation notation Any index that appears exactly twice in the same term is summed over
So we can also write the Lorentz transform as
$$ dr^\mu = \hat {\boldsymbol \Lambda^\mu_{\nu'}} dr^{\nu'} $$
Note that we always use superscripts for vector indices.
Define the spacetime metric
$$ \hat \eta = \begin{pmatrix} - 1& 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} $$
The square of the spacetime interval is
$$ ds^2 = \eta_{\mu \nu} dr^\mu dr^\nu $$
The spacetime interval is known as proper time when it is referring to the time duration in the particle’s frame, and proper distance when referring to the spatial distance when the two events are simultaneous. Just like the Cartesian distance in Galilean relativity, the spacetime interval between two events is invariant to Lorentz transforms.
When $ds^2>0$, the events are spacelike separated, and when $ds^2 < 0$ the events are timelike separated. When $ds^2=0$ the events are null separated.