Showing that Lorentz transformations are the only ones possible

[bob900]

In a book ("The special theory of relativity by David Bohm") that I'm reading, it says that if (x,y,z,t) are coordinates in frame A, and (x',y',z',t') are coordinates in frame B moving with v in realtion to A, if we have (for a spherical wavefront)

c^2t^2 - x^2 - y^2 - z^2 = 0

and we require that in frame B,

c^2t'^2 - x'^2 - y'^2 - z'^2 = 0

then it can be shown that the only possible transformations (x,y,z,t) -> (x',y',z',t') which leave the above relationship invariant are the Lorentz transformations (aside from rotations and reflections).

I'm wondering how exactly can this be shown?

 

[dextercioby]

To show it for a general Lorentz-Herglotz transformation is really difficult, you should only consider a (lorentzian) boost along Ox, for example, i.e. equal y to 0 and z to 0.

You should consider then x(x',t') and t(x',t') to be linear functions. Place some unknown coefficients and then determine them from physical assumptions.

 

[Mentz114]

Any transformation of the form
\left[<br /> \begin{matrix} a &amp; \sqrt{a^2-1}\\<br /> \sqrt{a^2-1} &amp; a<br /> \end{matrix} \right] \left[ \begin{matrix}dt\\<br /> dx<br /> \end{matrix} \right] = \left[ \begin{matrix}dt&#039;\\<br /> dx&#039;<br /> \end{matrix} \right]<br />
will preserve -dt'² + dx'² = -dt² + dx². More restraints than preserving the interval are needed.

 

[bob900]

Any transformation of the form
\left[<br /> \begin{matrix} a &amp; \sqrt{a^2-1}\\<br /> \sqrt{a^2-1} &amp; a<br /> \end{matrix} \right] \left[ \begin{matrix}dt\\<br /> dx<br /> \end{matrix} \right] = \left[ \begin{matrix}dt&#039;\\<br /> dx&#039;<br /> \end{matrix} \right]<br />
will preserve -dt'² + dx'² = -dt² + dx². More restraints than preserving the interval are needed.

Why wouldn't something as simple as :

x&#039; = x - k

t&#039; = \sqrt{t^2-2kx+k^2}

work (where k is some constant)?

Seems like that, and any other similarly arbitrary transformation could work...

 

[dextercioby]

[...]. More restraints than preserving the interval are needed.

Preserving the interval will ensure linearity of the transformations and only that.

 

[Mentz114]

Preserving the interval will ensure linearity of the transformations and only that.

Yes.
Taking the Taylor expansion of the matrix and dropping terms of order a² or greater gives the generator, I think. Exponentiating this gives a = cosh(something) but no idea what 'something' is. That's probably an illicit fudge, in any case.

 

[bob900]

To show it for a general Lorentz-Herglotz transformation is really difficult, you should only consider a (lorentzian) boost along Ox, for example, i.e. equal y to 0 and z to 0.

So if given just the following pieces of information :

1. c^2 t^2 - x^2 - y^2 - z^2 = 0
2. c^2 t&#039;^2 - x&#039;^2 - y&#039;^2 - z&#039;^2 = 0

is it "difficult" or actually impossible to show that the Lorentz transformation is the only possibility (aside from rotation x^2+y^2+z^2=x&#039;^2+y&#039;^2+z&#039;^2 and t=t', and reflection x=-x', t=-t', etc.)?

You should consider then x(x',t') and t(x',t') to be linear functions. Place some unknown coefficients and then determine them from physical assumptions.

That I know how to do - what I'm trying to see is if the book is wrong in saying that you only need 1 and 2 above. Here's a quote from the book :

The question then naturally arises as to whether there are any other transformations that leave the speed of light invariant. The answer is that if we make the physically reasonable requirement that the transformation possesses no singular points (so that it is everywhere regular and continuous) then it can be shown that the Lorentz transformations plus rotations plus reflections are the only ones that are possible.

 

[Fredrik]

I will use units such that c=1. I will also use the definition
$$\eta=\begin{pmatrix}-1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{pmatrix},$$ because I'm more used to this sign convention than the other one. The Minkowski form (pseudo-inner product) on $\mathbb R^4$ is defined by $g(x,x)=x^T\eta x$ for all $x\in\mathbb R^4$. Define $P=\{x\in\mathbb R^4|g(x,x)=0\}$. The OP is asking us to prove the following statement:

If $g(\Lambda(x),\Lambda(x))=g(x,x)$ for all $x\in P$, then $\Lambda$ is a Lorentz transformation.​

I don't know how to do that. I don't even know if it's possible. But I can prove a similar theorem that starts with a stronger assumption:

If $\Lambda$ is linear and $g(\Lambda x,\Lambda x)=g(x,x)$ for all $x\in\mathbb R^4$, then $\Lambda$ is a Lorentz transformation.​

Proof: Suppose that $\Lambda$ is linear and that $g(\Lambda x,\Lambda x)=g(x,x)$ for all $x\in\mathbb R^4$. Let $y,z\in\mathbb R^4$ be arbitrary. We have $$g(\Lambda(y-z),\Lambda(y-z))=g(y-z,y-z).$$ If we expand this using the linearity of $\Lambda$ and the bilinearity of g, and use that $g(\Lambda x,\Lambda x)=g(x,x)$ for all $x\in\mathbb R^4$, we see that $g(\Lambda y,\Lambda z)=g(y,z)$. Since y,z are arbitrary, this means that we have proved the following statement: For all $x,y\in\mathbb R^4$, $g(\Lambda x,\Lambda y)=g(x,y)$. Now let $x,y\in\mathbb R^4$ be arbitrary. We have $$x^T\eta y=g(x,y)=g(\Lambda x,\Lambda y)=x^T\Lambda^T\eta\Lambda y.$$ Let $\{e_\mu\}_{\mu=0}^3$ be the standard basis for $\mathbb R^4$. I will use the notation $M_{\mu\nu}$ for the component on row $\mu$, column $\nu$, of a matrix $M$. For all $\mu,\nu\in\{0,1,2,3\}$, we have
$$\eta_{\mu\nu}=e_\mu{}^T\eta e_\nu=e_\mu{}^T\Lambda^T\eta\Lambda e_\nu=(\Lambda^T\eta\Lambda)_{\mu\nu}.$$ So $\Lambda^T\eta\Lambda=\eta$, and this means that $\Lambda$ is a Lorentz transformation.

(The definition of "Lorentz transformation" goes like this: A linear $\Lambda:\mathbb R^4\to\mathbb R^4$ is said to be a Lorentz transformation if $\Lambda^T\eta\Lambda=\eta$).

 

[Fredrik]

Hm, it looks like I can also prove the following variant:

If $\Lambda$ is surjective, and $g(\Lambda(x),\Lambda(y))=g(x,y)$ for all $x,y\in\mathbb R^4$, then $\Lambda$ is a Lorentz transformation.​

With these assumptions, I can prove linearity by messing around with the expression
$$g(x,ay+bz)=g(\Lambda(x),\Lambda(ay+bz)).$$

 

[bcrowell]

Any transformation of the form
\left[<br /> \begin{matrix} a &amp; \sqrt{a^2-1}\\<br /> \sqrt{a^2-1} &amp; a<br /> \end{matrix} \right] \left[ \begin{matrix}dt\\<br /> dx<br /> \end{matrix} \right] = \left[ \begin{matrix}dt&#039;\\<br /> dx&#039;<br /> \end{matrix} \right]<br />
will preserve -dt'² + dx'² = -dt² + dx². More restraints than preserving the interval are needed.

Your a is just \gamma. I don't think there is any additional constraint needed, other than that the transformations should take the t axis to a line x=vt for some real v, and also not flip the orientation of the positive t axis. (These criteria rule out a<1).

 

[strangerep]

So if given just the following pieces of information :

1. c^2 t^2 - x^2 - y^2 - z^2 = 0
2. c^2 t&#039;^2 - x&#039;^2 - y&#039;^2 - z&#039;^2 = 0

is it "difficult" or actually impossible to show that the Lorentz transformation is the only possibility (aside from rotation x^2+y^2+z^2=x&#039;^2+y&#039;^2+z&#039;^2 and t=t', and reflection x=-x', t=-t', etc.)?

Lorentz transformations are not the only possibility. The most general transformations of this kind are the conformal transformations. There's an older thread over in the tutorials forum which derives them, but from the point of view of finding transformations that leave the metric invariant up to a scale factor.

Alternatively, it is possible to find the conformal transformations by direct solution of the differential equations defining the transformation. The (messy, difficult) details can be found in Appendix A of this older text:

V. Fock, N. Kemmer (translator),
The theory of space, time and gravitation.
2nd revised edition. Pergamon Press, Oxford, London, New York, Paris (1964).

You might be able to access a copy at Library Genesis. ;-)

Fock also shows that if you assume only the relativity principle (equivalence of inertial observers) then the most general transformations are of linear-fractional form -- which are not the same as conformal transformations since the latter involve a quadratic denominator in general. But if we add the light principle (which is what you used above), then the "intersection" between linear-fractional and conformal transformations is indeed the Lorentz transformations.

Regarding what the book said about requiring that the transformations be well behaved everywhere: although the more general transformations fail to be nonsingular in general everywhere, there have been recent attempts to use them to construct foundations that might account for the success of the Lambda-CDM model in cosmology -- the singular part of the transformation only occurs at the radius of the universe. But this is probably a subject for the BTSM forum.

 

[bob900]

So $\Lambda^T\eta\Lambda=\eta$, and this means that $\Lambda$ is a Lorentz transformation.

(The definition of "Lorentz transformation" goes like this: A linear $\Lambda:\mathbb R^4\to\mathbb R^4$ is said to be a Lorentz transformation if $\Lambda^T\eta\Lambda=\eta$).

But how is this definition of the Lorentz transformation equivalent to the "standard" definition:

x&#039; = \frac{x-vt}{\sqrt{1-v^2}}, t&#039; = \frac{t-vx}{\sqrt{1-v^2}}, y&#039;=y, z&#039;=z



?

 

[Mentz114]

Your a is just \gamma. I don't think there is any additional constraint needed, other than that the transformations should take the t axis to a line x=vt for some real v, and also not flip the orientation of the positive t axis. (These criteria rule out a<1).

No, a is not \gamma. It can be anything you like. The values a < 1 are ruled out because we want a real result. This transformation keeps the interval invariant. It's still a long way short of the LT.

Frederik's calculation shows it's not trivial to get the LT from a few assumptions.

 

[Fredrik]

But how is this definition of the Lorentz transformation equivalent to the "standard" definition:

x&#039; = \frac{x-vt}{\sqrt{1-v^2}}, t&#039; = \frac{t-vx}{\sqrt{1-v^2}}, y&#039;=y, z&#039;=z

This isn't the most general Lorentz transformation. This is just a boost in the x direction. A Lorentz transformation is a member of the Lorentz group, and the Lorentz group includes parity (=reversal of the spatial axes), time reversal, rotations and boosts (in arbitrary directions).

Let's consider the 1+1, dimensional case. If we write
$$\Lambda=\begin{pmatrix}a & b\\ c & d\end{pmatrix},$$ we can easily see that a≠0 and that c/a=-v. To see this, first note that
$$\Lambda\begin{pmatrix}1\\ 0\end{pmatrix}=\begin{pmatrix}a\\ c\end{pmatrix}.$$ In my notation, the upper component of a 2×1 matrix is the time coordinate. I will refer to it as "the 0 component", and the lower component, i.e. the spatial coordinate, as "the 1 component". I will also number the rows and columns of my 2×2 matrices from 0 to 1. For example, the 01 component of $\Lambda$ is b. If a=0, then the result above tells us that $\Lambda$ takes the time axis of the "old" coordinate system to the spatial axis of the "new" coordinate system. This corresponds to an infinite velocity difference, because the time axis is the (coordinate representation of) the world line of an object with velocity 0, and the spatial axis is the (coordinate representation of) the world line of an object with infinite velocity. This is why we can rule out a=0. This allows us to take a outside the coordinate matrix.
$$\Lambda\begin{pmatrix}1\\ 0\end{pmatrix}=\begin{pmatrix}a\\ c\end{pmatrix} =a\begin{pmatrix}1\\ c/a\end{pmatrix}.$$ Now we can interpret c/a as -v, because we know that $\Lambda$ maps the time axis to the line
$$t\mapsto t\begin{pmatrix}1 \\ -v\end{pmatrix},$$ i.e. the line with x'=-vt'.

Now consider the effect of $\Lambda$ on two coordinate matrices $\begin{pmatrix}t_1\\ 0\end{pmatrix}$ and $\begin{pmatrix}t_2\\ 0\end{pmatrix}$ with $t_1<t_2$.
\begin{align}
\Lambda\begin{pmatrix}t_1\\ 0\end{pmatrix} =\begin{pmatrix}at_1\\ ct_1\end{pmatrix},\qquad
\Lambda\begin{pmatrix}t_2\\ 0\end{pmatrix} =\begin{pmatrix}at_2\\ ct_2\end{pmatrix}\end{align} The 0 components of the new coordinate pairs are $at_1$ and $at_2$ respectively. If a>0, then $t_1<t_2$ implies that $at_1<at_2$, but if a<0, then $t_1<t_2$ implies that $at_1>at_2$. So $\Lambda$ preserves the temporal order of events on the 0 axis when a>0, and reverses them when a<0. To get the specific result you want, we need to assume that a>0. We are now dealing with an orthochronous Lorentz transformation.

A similar argument shows that $\Lambda$ preserves the order of events on the spatial axis when d>0 and reverses them when d<0. So we also assume that d>0. We are now dealing with a proper Lorentz transformation. A Lorentz transformation that's both proper and orthochronous is sometimes called a restricted Lorentz transformation.

Because of the above, we will write $\Lambda$ as
$$\Lambda=\gamma\begin{pmatrix}1 & \alpha\\ -v & \beta\end{pmatrix},$$ where $\gamma,\beta>0$.
\begin{align}
\begin{pmatrix}-1 & 0\\ 0 & 1\end{pmatrix} &=\eta =\Lambda^T\eta\Lambda =\gamma^2\begin{pmatrix}1 & -v\\ \alpha & \beta\end{pmatrix}\begin{pmatrix}-1 & 0\\ 0 & 1\end{pmatrix}\begin{pmatrix}1 & \alpha\\ -v & \beta\end{pmatrix}\\
&=\gamma^2\begin{pmatrix}1 & -v\\ \alpha & \beta\end{pmatrix}\begin{pmatrix}-1 & -\alpha\\ -v & \beta\end{pmatrix} =\gamma^2\begin{pmatrix}-1+v^2 & -\alpha-v\beta\\ -\alpha-v\beta & -\alpha^2+\beta^2\end{pmatrix}
\end{align}
The 00 component of this equality tells us that $-1=\gamma^2(-1+v^2)$, which implies both that |v|<1 (because $\gamma^2>0$) and that $$\gamma=\frac{1}{\sqrt{1-v^2}}.$$ The 01 and 10 components both tell us that $\gamma^2(-\alpha-v\beta)=0$, which implies that $\alpha=-v\beta$. The 11 component tells us that $\gamma^2(-\alpha^2+\beta^2)=1$. So
\begin{align}1-v^2 &=\frac{1}{\gamma^2} =-\alpha^2+\beta^2 =-\beta^2v^2+\beta^2=\beta^2(1-v^2)\\
\beta &=1\\
\alpha &= -v\beta=-v.
\end{align} So our final result for $\Lambda$ is
$$\Lambda=\frac{1}{\sqrt{1-v^2}}\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix}.$$
If you prefer to write this out as a system of equations,
$$\begin{pmatrix}t'\\ x'\end{pmatrix}=\gamma\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix}\begin{pmatrix}t\\ x\end{pmatrix}=\gamma\begin{pmatrix}t-vx\\ -vt+x\end{pmatrix}$$ \begin{align}
t' &=\gamma(t-vx)\\
x' &=\gamma(x-vt).
\end{align}

 

[bcrowell]

No, a is not \gamma. It can be anything you like. The values a < 1 are ruled out because we want a real result. This transformation keeps the interval invariant. It's still a long way short of the LT.

Frederik's calculation shows it's not trivial to get the LT from a few assumptions.

No, a is not simply anything you like. It's gamma. You can tell it's gamma because of the transformation's action on the t axis, which slants it with a slope v. Fredrik's calculation is unnecessarily complicated.

 

[Mentz114]

No, a is not simply anything you like. It's gamma. You can tell it's gamma because of the transformation's action on the t axis, which slants it with a slope v. Fredrik's calculation is unnecessarily complicated.

Even if a were 1/v (say) the proper length would be invariant. a=γ and a=cosh(R) are two special cases. I'm addressing the question whether preserving the proper interval is sufficient to get the LT - and I'm asserting it is not.

 

[bcrowell]

Even if a were 1/v (say) the proper length would be invariant. a=γ and a=cosh(R) are two special cases. I'm addressing the question whether preserving the proper interval is sufficient to get the LT - and I'm asserting it is not.

You can't have a be anything but gamma, because v is defined by the action of the LT on the positive t axis.

No, a is not \gamma.The values a < 1 are ruled out because we want a real result.

Actually this only rules out -1 \lt a lt 1. What rules out all values of a<1 is the definition of v.

 

[Erland]

It is always assumed that the transformation is linear (at least if the origin is mapped to the origin, otherwise affine). But what is the physical reason for this assumption?

 

[Mentz114]

You can't have a be anything but gamma, because v is defined by the action of the LT on the positive t axis.
..
..
Actually this only rules out -1 \lt a lt 1. What rules out all values of a<1 is the definition of v.

Ben, I think we're talking across each other so I'll let it go now.

 

[Fredrik]

Fredrik's calculation is unnecessarily complicated.

In what way? What part of it can be simplified?

 

[Fredrik]

It is always assumed that the transformation is linear (at least if the origin is mapped to the origin, otherwise affine). But what is the physical reason for this assumption?

The idea is that for each inertial (=non-accelerating) observer, there's a coordinate system in which the observer's own motion is described by the time axis, and the motion of any non-accelerating object is described by a straight line. So a function that changes coordinates from one of these coordinate systems to another must take straight lines to straight lines.

 

[Erland]

The idea is that for each inertial (=non-accelerating) observer, there's a coordinate system in which the observer's own motion is described by the time axis, and the motion of any non-accelerating object is described by a straight line. So a function that changes coordinates from one of these coordinate systems to another must take straight lines to straight lines.

Hmm, are you saying something like that a map between vector spaces that takes lines to lines must be linear, or affine?
Well, that's certainly not true in one dimension, where the map f(x)=x^3 maps the entire line onto itself without being linear, or affine.
But perhaps in higher dimensions...? Is there a theorem of this kind?

 

[samalkhaiat]

In a book ("The special theory of relativity by David Bohm") that I'm reading, it says that if (x,y,z,t) are coordinates in frame A, and (x',y',z',t') are coordinates in frame B moving with v in realtion to A, if we have (for a spherical wavefront)

c^2t^2 - x^2 - y^2 - z^2 = 0

and we require that in frame B,

c^2t&#039;^2 - x&#039;^2 - y&#039;^2 - z&#039;^2 = 0

then it can be shown that the only possible transformations (x,y,z,t) -> (x',y',z',t') which leave the above relationship invariant are the Lorentz transformations (aside from rotations and reflections).

I'm wondering how exactly can this be shown?

I don't think Bohm said this! Lorentz group is a subgroup of a bigger group called the conformal group. It is the conformal group that preserves the light-cone structur.

Sam

 

[pervect]

Didn't the paper that Ben mentioned in another thread, http://arxiv.org/abs/physics/0302045, go through all this?

The assumptions that that paper made were (skimming)

* replacing v with -v must invert the transform
* isotropy
*homogeneity of space and time

with a few tricks along the way:
* adding a third frame
* noting that x=vt implies x'=0

The result was pretty much that there must be some invariant velocity that was the same for all observers. (THere were some arguments about sign of a constant before this to establish that it was positive). The remaining step is to identify this with the speed of light.

 

[samalkhaiat]

So if given just the following pieces of information :

1. c^2 t^2 - x^2 - y^2 - z^2 = 0
2. c^2 t&#039;^2 - x&#039;^2 - y&#039;^2 - z&#039;^2 = 0

is it "difficult" or actually impossible to show that the Lorentz transformation is the only possibility (aside from rotation x^2+y^2+z^2=x&#039;^2+y&#039;^2+z&#039;^2 and t=t', and reflection x=-x', t=-t', etc.)?
That I know how to do - what I'm trying to see is if the book is wrong in saying that you only need 1 and 2 above. Here's a quote from the book :

​

Now Bohm is making sense.

see post #9 in
www.physicsforums.com/showthread.php?t=420204

 

[strangerep]

It is always assumed that the transformation is linear (at least if the origin is mapped to the origin, otherwise affine). But what is the physical reason for this assumption?

The most common reason is so-called homogeneity of space and time. By this, the authors mean that position-dependent (and time-dependent) dilations (scale changes) are ruled out arbitrarily.

Personally, I prefer a different definition of spacetime homogeneity: i.e., that it should look the same wherever and whenever you are. IOW, it must be a space of constant curvature.
This includes such things as deSitter spacetime, and admits a larger class of possibilities.

But another way that various authors reach the linearity assumption is to start with the most general transformations preserving inertial motion, which are fractional-linear transformations. (These are the most general transformations which map straight lines to straight lines -- see note #1.) They then demand that the transformations must be well-defined everywhere, which forces the denominator in the FL transformations to be restricted to a constant, leaving us with affine transformations.

In the light of modern cosmology, these arbitrary restrictions are becoming questionable.

--------
Note #1: a simpler version of Fock's proof can be found in Appendix B of this paper:
http://arxiv.org/abs/gr-qc/0703078c/0703078 by Guo et al.

An even simpler proof for the case of 1+1D can also be found in Appendix 1 of this paper:
http://arxiv.org/abs/physics/9909009 by Stepanov. (Take the main body of this paper with a large grain of salt, but his Appendix 1 seems to be ok, though it still needs the reader to fill in some of the steps -- speaking from personal experience. :-)

 

[Fredrik]

Hmm, are you saying something like that a map between vector spaces that takes lines to lines must be linear, or affine?
Well, that's certainly not true in one dimension, where the map f(x)=x^3 maps the entire line onto itself without being linear, or affine.
But perhaps in higher dimensions...? Is there a theorem of this kind?

The only book I know that suggests that there is such a theorem left the proof as an exercise. I tried to prove it a couple of years ago, but got stuck and put it aside. I just tried again, and I still don't see how to do it. It's pretty annoying. Three distinct vectors x,y,z are said to be collinear if they're on the same straight line. So x,y,z are collinear if and only if they're all different and there's a number a such that $z=x+a(y-x)$, right? Note that the right-hand side is $=(1-a)x+ay$. So three vectors are collinear if and only if they're all different and (any) one of them can be expressed as this special type of linear combination of the other two.

A linear transformation $T:U\to V$ is said to preserve collinearity if for all collinear x,y,z in U, Tx,Ty,Tz are collinear.

It's trivial to prove that linear maps preserve collinearity. Since $T(ax+by)=aTx+bTy$ for all a,b, we have $T((1-a)x+ay)=(1-a)Tx+aTy$ for all a.

I still haven't been able to prove that if T preserves collinearity, T is linear. Suppose that T preserves collinearity. Let x,y be arbitrary vectors and a,b arbitrary numbers. One idea I had was to rewrite $T(ax+by)=T(ax+(1-a)z)$. All I have to do is to define $z=by/(1-a)$. But this is a lot less rewarding than I hoped. All we can say now is that there's a number c such that
$$T(ax+by)=cTx+(1-c)Tz =cTx+(1-c)T\left(\frac{by}{1-a}\right).$$ The fact that we can't even carry the numbers a,b over to the right-hand side is especially troubling. I don't know, maybe I've misunderstood a definition or something.

The book I'm talking about is "Functional analysis: Spectral theory" by Sunder. It can be downloaded legally from the author's web page. Scroll down to the first horizontal line to find the download link. See exercise 1.3.1 (2) on page 9 (in the pdf, it may be on another page in the actual book). Edit: Direct link to the pdf.

 

[strangerep]

The only book I know that suggests that there is such a theorem left the proof as an exercise. I tried to prove it a couple of years ago, but got stuck and put it aside. I just tried again, and I still don't see how to do it. It's pretty annoying. [...]

I guess you didn't see my previous post #27, huh? :-)

 

[Fredrik]

I guess you didn't see my previous post #27, huh? :-)

Not until after I posted. I'm checking out those appendices now. I guess Sunder's exercise is just wrong then. No wonder I found it so hard to solve it.

 

[strangerep]

I guess Sunder's exercise is just wrong then.

He's restricting himself to the case of a vector space and linear transformations between them. But the more general case involves differentiable coordinate transformations on a more general manifold -- which is a different problem.

Edit: looking at his exercise, I think he means "$x,y,z$ in $V$", meaning that $x,y,z$ are vectors in $V$. So the "straight line" also includes the origin. That makes his exercise almost trivial because "being on a straight line" means that the vectors are all simple multiples of each other (i.e., they're on the same ray), and linear transformations preserve this.

But this is somewhat tangential to the current issue since in relativity we want something more general which preserves continuous inertial motion.