# Why is this translation not a linear transformation? [closed]

From Linear map, the sixth example:

The translation $$x \rightarrow x+1$$ is not a linear transformation. Why?

What about $$x \rightarrow x +dx$$? Is this translation a linear transformation?

Does it matter if the transformation is not linear?

• About x+dx could you be more precise? I dont think it is a well defined transformation at all. Like lets take an example: i want to transform (1) (lets assume 1d space). Which point would be the result of adding dx? Commented Nov 26, 2019 at 6:43
• Since many students seriously encounter Linear Algebra for the first time when they study Quantum Mechanics, it's worth pointing out that the translation operator is indeed linear and its generator is the momentum operator which in the position representation takes the form $-i\hbar\frac{d}{dx}$. So how can those be linear in spite of the example you found on Wikipedia? The QM operators don't operate on the underlying space, they operate on the states, complex-valued functions of $x$ in the 1D position representation. You don't replace $x\to x+1$, you replace $\psi(x)\to\psi(x-1)$. Commented Nov 26, 2019 at 13:24
• Because the transformation doesn’t satisfy the rules for a linear operator. In particular, it must send the zero to zero, which isn’t the case. Commented Nov 27, 2019 at 16:12
• A clarification since this moved to math.stackechange: When I wrote the above comment, I was referring to physics students since this question was still on physics.stackexchange. I still think that this question is more likely to arise for budding physicists than developing mathematicians. Commented Nov 29, 2019 at 3:29
• For one thing the transformation x->x+1 doesn't map 0 to 0. I don't know what you mean by "dx" in "1+ dx". Commented Aug 27, 2021 at 20:30

OP's transformations are affine transformations. Whether they are called linear transformations depends on context and conventions.

• Within the context of linear algebra, a linear transformation maps the zero vector into the zero vector. Then OP's transformations are generically not linear.

• In other contexts/conventions, linear & affine transformations are the same thing.

• +1: I am not sure about mathematical terminology but in physics, in practice, the two are distinguished (if needed) by specifying as to whether we are talking about homogeneous linear transformations or inhomogeneous linear transformations.
– user359328
Commented Nov 25, 2019 at 12:12
• @Dvij Mankad: Good point. Commented Nov 25, 2019 at 12:15
• One could mention that affine transformations are 'more physical'than linear ones (since there is no distinct origin in space) Commented Nov 26, 2019 at 6:41
• Yes, and in some contexts/conventions, it's a proper subset. Commented Nov 26, 2019 at 8:47
• @lalala It depends a bit one what you transform. If you transform space, where there is no origin, then sure (though you would want it to also be orthogonal). If you transform wave functions in quantum mechanics, then linear (and unitary) is the 'more physical' thing. But to agree with your point, it always pays well to keep in mind, which operations are natural on the object you are working on and which aren't.
– mlk
Commented Nov 26, 2019 at 10:18

The definition of a linear transformation is: $$T(u+v) = T(u)+T(v)$$, (and also $$T(au)=aT(u)$$).

The translation $$x\rightarrow x+1$$ is not linear transformation. Why?

In your case $$T(u)=u+1$$ and $$T(v)=v+1$$. So $$T(u)+T(v)=u+v+2$$. Whereas $$T(u+v)=u+v+1$$

So $$T(u+v) = T(u)+T(v)$$ does not hold true and T is not linear.

What about $$x\rightarrow x+dx$$, is this translation a linear tranformation?

In the case of $$x \rightarrow x +dx$$; $$x+y \rightarrow x+y +dx +dy = (x+dx) +(y+dy)$$ so this transformation is linear.

Does it matter if the transformation is not linear?

It's super important: Most engineering and physics analysis depends on linearity. Non linear analysis is a field of study by itself. For example the Fourier Transform depends on linearity.

Translation in an $$n$$-dimensional space is not a linear operation but you can make it a linear operation by looking at it from another space. The price of this is adding another dimension. So translation in $$n$$ dimensions can be expressed as a linear operation in $$n+1$$ dimensions. In the case of one dimension it looks like this: you identify the point $$x$$ with the point $$\binom{x}{1}$$ in the two-dimensional space. The key is that the second coordinate be equal to one. Let's denote this identification as $$x\doteq \binom{x}{1}$$. Translation $$x\to x+a$$ can then be expressed as $$x\to x+a \doteq \begin{pmatrix}x + a\\ 1\end{pmatrix} = \begin{pmatrix}1 & a \\ 0 & 1\end{pmatrix}\begin{pmatrix}x\\ 1\end{pmatrix},$$ which is matrix multiplication. Now, on this two-dimensional space, translation by $$a$$ along the direction of the one-dimensional space is indeed a linear operation and it can be expressed by the matrix $$(\begin{smallmatrix}1 & a\\0 & 1\end{smallmatrix})$$. It's easy to see how subsequent translations do the right thing and how their matrices combine. If you've never done it, it is also a fun exercise to extend this procedure to higher dimensions. Taking this line of thought further leads to topics such as Projective Spaces and the Affine Transformations mentioned in the previous answers.

Since PM 2Ring mentioned homogeneous coordinates in the comments, I thought it's perhaps worth pointing out that linearity in the two-dimensional space isn't what you might expect: e.g. $$2\times (\begin{smallmatrix}x\\ 1\end{smallmatrix})=(\begin{smallmatrix} 2x \\ 2\end{smallmatrix})$$ no longer has one as its second coordinate, and so we can no longer use the above prescription to identify it with a point in the one-dimensional space. This is to say that while we can identify a linear operation corresponding to translation, we need additional structure (which I won't discuss in this answer) to make sense of the linearity in terms of the original space.

• Nor to mention en.wikipedia.org/wiki/Homogeneous_coordinates Commented Nov 26, 2019 at 9:43
• Actually, since this is physics.stackexchange, I thought I'd point out that thinking about units is a fairly straightforward way to understand what homogeneous coordinates will do. I.e., where would you put meters, inches, feet, furlongs in the equations and how would that affect your thinking about linearity? Commented Nov 27, 2019 at 1:07
• So if a 3D translation expressed in 4D homogeneous coordinates becomes linear, shouldn't the transformation matrix be orthogonal? What am I missing? Commented Aug 27, 2021 at 20:09
• Orthogonal transformations are precisely the linear transformations that keep the euclidean length constant. You can easily check that $||\mathbf{x}||^2=||\mathbf{Ox}||^2\forall x$ implies $\mathbf{O^TO}=\mathbb{1}$ and vice versa. Why should translations preserve the lengths of vectors i.e. be orthogonal? Commented Aug 29, 2021 at 2:12