Why is the relative velocity of B with respect to A negative of the relative velocity of A with respect to B? I'm trying to figure out how to derive the transformation matrix for the Lorentz boost. Consider two inertial frames A and B, and let B move at a constant velocity V with respect to A. All the derivations I've seen assume that this implies that A moves at a velocity -V with respect to B. This does not seem obvious to me, can someone explain why this is true?
Edit : To clarify, I understand "inuitively" why this is true. What I'm interested in doing is deriving the Lorentz boost formula axiomatically - i.e., using the invariance of the speed of light as my starting point and showing that the Lorentz boost is the only linear transformation that satisfies this. What I don't get is how this reversal of velocity follows from this.
 A: Intuitively, you are comparing the velocities of two frames.  If there were unit masses moving in those frames, then there would be a center of mass frame.  That center of mass frame of the fictitious masses exists regardless of whether there are masses there.  And this frame sees each of the other frames moving in equal and opposite directions.  This frame sees them exchange radar ranging data back and for (round trip time by wristwatch divided by two times the speed of light a distance, compute that twice at different times, get a velocity).  And this third frame sees that the distance between the two frames grows at a steady rate and that their clocks are ticking at a steady rate.  The third frame thinks their clock rates are off, but off by the same amount.  So the third frame computes they find each other moving at the same relative velocity.
I started writing my answer before the question edit, so ...
**Original answer follows*
There is a geometrical view of special relativity where you imagine it spacetime as a 4D vector space with an interesting way to take two vectors and get a number, that only depends on the vectors, specifically the square of a vector $R=(w,x,y,z)$  is $R^2=w^2-x^2-y^2-z^2$.
Now that we know how to square vectors, we can define a scalar product (if we want $(R+S)^2=(R+S)*(R+S)=R*R+R*S+S*R+S*S$ and for the scalar product of vectors to be symmetric then) we can define $R*S=((R+S)^2-S^2-R^2)/2$, so now we note that vectors $W=(1,0,0,0)$, $X=(0,1,0,0)$, $Y=(0,0,1,0)$, and $Z=(0,0,0,1)$ are all mutually orthogonal.  So it's just like regular space except the sign of $W^2$ is the negative of the sign of the other basis vectors.
So we have a 4D vector space with an interesting way to take scalar products, but what about orthogonal vectors (for instance $X$ and $Y$), how do they multiply? Their scalar product X*Y is zero because that's what it means to be orthogonal.  And so 
$(X+Y)^2=(X+Y)*(X+Y)=X*X+2X*Y+Y^2=X^2+Y^2$
But just like we wanted $(R+S)^2=(R+S)*(R+S)=R*R+R*S+S*R+S*S$ for scalar products we can also make another product that has
$(R+S)(R+S)=R(R+S)+S(R+S)=RR+RS+SR+SS$, but like matrices the order we multiply matters for this product.
So for orthogonal vectors like $X$ and $Y$ we have $(X+Y)^2=X^2+Y^2$, and we generally have $(X+Y)^2=XX+XY+YX+YY$, so we have that for orthogonal vectors like $X$ and $Y$, we know that $XY+YX=0$, so for orthogonal vectors if you change the order you multiply them you get a minus sign.  (You might also notice that $RS+SR=2R*S$ so this is really the same as being orthogonal).
This is useful because we can then have easy ways to extract part of a vector that is orthogonal to another vector.  For instance if we have $S=aW+bX+cY+dZ$, then we can compute $WS=aWW+bWX+cWY+dWZ$ and $SW=aWW+bXW+cYW+dZW$ we can use the fact that orthogonal vectors multiplied in opposite orders give a minus sign to rewrite $SW$ as $SW=aWW-bWX-cWY-dWZ$ and if we subtract that from $WS$ we get $WS-SW=2(0WW+bWX+cWY+dWZ)$. Since $W^2=1$, we can multiply by $W$ (on the left) and divide by two, and get $(S-WSW)/2=bX+cY+dZ$, so we get just the part of the vector $S$ that was orthogonal to $W$ (If we'd have used a vector with a negative square, then we'd have had to divide by -1 after we had all the $W^2$ parts).
That's all the geometry and algebra we need for this problem to fully look at it the way this geometry handle motion. So let's get to the physics.
Now we can think of $w$ as $ct$ so this different direction (it has a different square by a factor of -1) is just time measured in units of distance.  So there is an observer that sees the vector $R$ as just spelling out how long they waited and how far they moved in the $X$, $Y$ and $Z$ directions.  So when they look at someone else's motion they look at how far away the point travelled and how long it took. So their spacetime displacement is $(c\Delta t, \Delta x, \Delta y, \Delta z)$ for the displacement, then divide by $\Delta t$ to get $(c,v_x,v_y,v_z)$, which is just a rescaled vector with units of velocity that goes from its spacetime location at one time to its spacetime location at another time.  So the velocity you see someone else move is just the part of their actual motion vector $M=(c\Delta t, \Delta x, \Delta y, \Delta z)$ that is orthogonal to your $W$ vector i.e. $(0, \Delta x, \Delta y, \Delta z)$ divided by the time component of their motion vector $c\Delta t$ all multiplied by $c$.  So $V=c\frac{1}{c\Delta t}(0, \Delta x, \Delta y, \Delta z)=(0,v_x,v_y,v_z)$.  But theses are all things we computed in the geometry and algebra section so $V=c\frac{(M-WMW)/2}{(MW+WM)/2)}$  and in fact since $W$ is so easy to eliminate you might even prefer the object $VW=c\frac{MW-WM}{MW+WM}$.
This is something easy to compute, and the RHS $c\frac{MW-WM}{MW+WM}$ doesn't even depend on how big $W$ is, you can use any vector that points in your time direction.  And it also doesn't depend on how big $M$ is, any vector pointing in the correct direction works.  Geometrically that top factor is like a $sin(\theta)$ that also keeps track of the plane that $V$ and $W$ are in and the bottom is like a $\cos(\theta)$, and the ratio makes it not matter how big the vectors are, so this quantity $VW$ is like a measure of the spacetime angle between $M$ and $W$, in a tangent type form.
So let's do relativity!  Relativity says that your time and space basis vectors are just as good as anybody else's.  So when that person measures your speed he uses your $W$ for $M$ (it points in the direction you are going $(1,0,0,0)$, and he can use his own motion for $W$.  So he computes $c\frac{WM-MW}{WM+MW}$ for velocity times his unit vector $M/\sqrt{M^2}$, and they are negatives of each other because the tops are negatives and the bottoms are the same.
This technique actually works to break down every geometrical object into the parts different observers see.  For instance if you see an electric and magnetic field you have six numbers, $E_x,E_y,E_z,B_x,B_y,$ and $B_z$ which actually are the components of one six component object (the electromagnetic field $F$) built out of the basis object $WX, WY, WZ, YZ, XZ,$ and $XY$.  So to see what electric and magnetic parts someone else sees you can take their unit motion vector $U=M/\sqrt{M^2}$ compute $FU$ and $UF$ and add them $FU+UF$ get the three components for the magnetic field $B=(UF+UF)/2$ and subtract them to get the three components of the electric field $E=(UF-FU)/2$.  So there is one electromagnetic field and it's a simple multiplication to find out how people see the parts.  Everyone can make the same geometrical objects and break them down the same way.
Also this same math allows you to talk about rotations and planes in 4D (there isn't a vector orthogonal to a plane anymore) and the price you have to pay is you have to pay attention to the order you multiply, but you are used to that from rotations, and if you practice this then you already have tools you need to do relativistic quantum mechanics too.
edit to show how to get relative velocity
If you have one person moving at $\frac{3}{5}c$ in the x direction and someone moving at $\frac{12}{13}c$ in the x direction then you can easily see what relative velcity they see for each other.  One moves in the direction $A=(5,3,0,0)$ and the other one moves in the direction $B=(13,12,0,0)$ so we compute $c\frac{AB-BA}{AB+BA}$
$=c\frac{(5W13W+5W12X+3X13W+3X12X)-(13W5W+12X5W+13WX3+12X132X)}{(5W13W+5W12X+3X13W+3X12X)+(13W5W+12X5W+13WX3+12X132X)}$
$=c\frac{5W12X+3X13W}{(5W13W+3X12X)}=c\frac{(3*13-5*12)XW}{(5*13-3*12)}=cXW\frac{(3*13-5*12)}{(5*13-3*12)}.$
Which gives the direction ($XW$ is the plane of relative velocity, any yes relative velocity is really a plane in spacetime everyone agrees on the plane two motions span and the angle between them, just multiply by the unit motion vector of one of the two if you want the spatial vector direction they see) and the magnitude (just like the velocity addition formula).
The two motions could be in any directions and the same formula works, it's just about seeing which part of a motion one observer sees as orthogonal to themselves.
A: A snappy but unsatisfying answer is that relative velocity just means the difference in velocity, and difference denotes subtraction. And x - y is the opposite of y - x; in other words if z = x - y, then y - x = -z.
A picture that might help is this: You are on train A, moving north at 40 m/s. On a parallel track is train B, moving north at 50 m/s. So v_BA, the relative velocity of B as observed from A, is 10 m/s north. In other words, you see the other train passing you at 10 m/s, as you both travel north.
A passenger on the other train, of course, sees your train slipping behind at a rate of 10 m/s. So your relative speed as observed from her train, v_AB, is 10 m/s south, or 10 m/s in the opposite direction.
This clearly generalizes, so we can say that v_AB = -v_BA for any two frames.
No fancy SR stuff required; this is true for Galilean relativity or really any two vector quantities in inertial frames.
