I don't know whether I can ask this question here, but I am asking here, I couldn't find any correct explanation about my doubts.

I watched a youtube video about geometrical idea of a Linear Transformation from $\mathbb{R^2}$ to $\mathbb{R^2}$, it said, a transformation is said to be linear if the grids are equally spaced after the transformation, and the zero should remain its position after the transformation.

But the I find tough to relate this idea to the definition of the linear transformation

i.e $T(x+y)=T(x)+T(y)$ and $T(\alpha x)=\alpha T(x)$

Can anyone give example for the Transformation which satisfies the first condition but not second and a transformation which satisfies second condition but not the first one

  • $\begingroup$ See Linear map. $\endgroup$ – Mauro ALLEGRANZA Sep 11 at 14:15
  • $\begingroup$ Examples that I am looking for? $\endgroup$ – Madhan Kumar Sep 11 at 14:18
  • $\begingroup$ @MauroALLEGRANZA -- The OP asked for an example of an additive, non linear function, and an example for a homogeneous, non additive function. Where exactly in your wikipedia page can he find such examples? $\endgroup$ – uniquesolution Sep 11 at 14:24
  • $\begingroup$ Regarding additive but not homogeneous transformations: math.stackexchange.com/questions/1836862/…. $\endgroup$ – Hans Lundmark Sep 11 at 14:41
  • $\begingroup$ The characterization in terms of grids is only valid for nonsingular transformations. A linear transformation might also collapse the plane to a single line or even a point. I suppose that you could consider the grids to still be “evenly spaced” in these cases, too. $\endgroup$ – amd Sep 11 at 20:46

It is easier to understand the geometry of linear maps in terms of matrices. Let $\mathbb{R}^2$ be equipped with its canonical base. Then all linear maps $\varphi : \mathbb{R}^2 \to \mathbb{R}^2$ can be written as

$$ \varphi(x) = Ax $$

for some suitable, real $2 \times 2$ -matrix $A$ with respect to the canonical (or any other) basis. This is a standard (but not completely trivial) result from linear algebra and follows from your definition of linearity.

So now you can "forget" about your abstract definition of linearity and think of linear maps in terms of matrix multiplications.

Your claim, that the "grids" of $\mathbb{R}^2$ are equally spaced after matrix multiplication is not true in general. This only applies in the case of orthonormal matrices (which correspond to compositions of rotations around the origin and isometric reflection-operations).

Generally speaking, a multiplication of a vector $(x,y)$ by a matrix $A$ (if it is regular, or, in other words, if its determinant does not vanish) can be seen as a coordinate transformation from the standard base to the base, which is formed by the two column vectors of the inverse $A^{-1}$ of $A$. Geometrically, this can do several things to $\mathbb{R}^2$, including stretching the space, rotating the space and reflecting the space on a subspace.

To really get a geometric feeling for linear maps, it is absolutely essential to study several examples in depth. For example, which matrix would rotate $\mathbb{R}^2$ counterclockwise with angle $\theta$? And which matrix would relfect every vector on the origin? Which matrix would distort the grid of $\mathbb{R}^2$ (there are infinitely many)?


In a vector space $V$ you have the sum $x+y$ and the scaling $\alpha x$. Therefore equations of the form $$z= x+ y,\qquad v=\alpha\, u\tag{1}$$ make sense in $V$. A map $T:\>V\to W$ between vector spaces is linear if truths of the form $(1)$ involving vectors and scalars are kept by $T$, i.e., if whenever $(1)$ holds we also have $$T(z)=T(x)+T(y),\qquad T(v)=\alpha\, T(u)\ .$$ The usual formulation of the axioms reduces the number of variables by writing $$T(x+y)=T(x)+T(y),\qquad T(\alpha\,u)=\alpha\,T(u)\ .\tag{2}$$

It is difficult to produce a $T$ satisfying $(2_1)$ but not $(2_2)$, since you would need the axiom of choice for this. A map $T$ satisfying $(2_2)$, but not $(2_1)$, you can obtain as follows: Consider an arbitrary odd function $f:\>S^{n-1}\to{\mathbb R}$ on the unit sphere of ${\mathbb R}^n$, i.e., $f(-x)=-f(x)$. Using this $f$ define $$T(x):=|x|\,f\left({x\over|x|}\right)\qquad\bigl(x\in{\mathbb R}^n\bigr)\ .$$


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