I have studied that given vector spaces $V_1$ and $V_2$, a function $T:V_1 \rightarrow V_2$ is called a linear transformation of $V_1$ into $V_2$, if following two properties are true for all $u, v \in V_1$ and scalar $c$:

$(1)$: $T(u+v) = T(u) + T(v)$ and $(2): T(cu) = c T(u) $.

My questions are

$1$: What is the geometrical interpretation of properties $1$ and $2$ which says that $T$ preserves additivity and scalar multiplication. I am not able to see this geometrically. What is the meaning of preserving additivity and scalar multiplication.

$2$: At some place I have studied that a linear transformation will be linear if it sends each line to line and planes to planes and so on. How can we interpret this based on these two properties.

I need help to understand this.

Thank you very much for your time..

  • 1
    $\begingroup$ Think of $T$ as a matrix is helpful. $\endgroup$
    – Yes
    Aug 27, 2014 at 4:42
  • 1
    $\begingroup$ For example, we can stretch a vector by some scalar and then rotate it, or we can rotate the vector and then stretch it. These two operations will produce the same vector, and are an example of how linear transformations preserve scalar multiplication. $\endgroup$
    – Nathan
    Aug 27, 2014 at 5:11
  • $\begingroup$ Consider $ k-$vector spaces $ V, W $. We can ask ourselves : What are the maps $ f : V \to W $ such that "equations in $ V $ give corresponding equations between images in $ W $" i.e. "$ \sum_{j=1}^{n} \lambda_j v_j = 0 $ in $ V $ implies $ \sum_{j=1}^{n}\lambda_j f(v_j) = 0 $ in $ W $" ? We see these are precisely the linear maps from $ V $ to $ W $. $\endgroup$ Sep 13, 2021 at 14:02

2 Answers 2


For Question 1,

The geometric interpretation largely has to do with the fact that if you have a vector $w$ that can be written as a linear combination of vectors $u$ and $v$ thus $w=u+v$ than if to see what happens to when we apply $T$ to $w$ (i.e. $T(w)$) we can think of this as what happens when we apply to $u$ and $v$ then just add those resulting vectors. Similarly the preservation of scalar multiplication is just saying we can see what happens when we apply $T$ to $u$ then scale it.

To really understand this, lets look at example. Let $T: \mathbb{R}^{2}\rightarrow \mathbb{R}^{2}$ that reflects vector about y axis. Lets look at what this does would be for vector $(1,2)$, well first note that $(1,2)=(1,0)+2(0,1)$ thus when looking at $T[(1,2)]$ we can see that because property additivity and scalar multiplication that we have $$T[(1,2)]=T[(1,0)]+2T[(0,1)]$$ Well when we reflect $(1,0)$ and $(0,1)$ about the y axis we get $(-1,0)$ and $(0,1)$ respectively thus we have $$=(-1,0)+2(0,1)=(-1,2)$$ Thus we have $T[(1,2)]=(-1,2)$. Well what are we doing geometrically? Well what we did was we broke up $(1,2)$ addition of its x direction and y direction. We then scaled these down to make them unit vectors, we then flipped each one about the y axis, then rescaled them back, then added them back together to get our resulting vector $(-1,2)$.

Now as you may know that $(0,1)$ and $(1,0)$ is basis for vectors in $\mathbb{R}^{2}$. Well whenever we have have maps that have properties of linear maps this means we can always just find what the resulting vector is from applying a linear map to that vector, by just thinking of it as simply the method of scaling down vector to sum of basis vectors, transforming those basis vectors, then rescaling and adding these resulting vectors.

Im sorry if this answer isn't very good, and take it with a grain of salt, this is just how Ive always thought of it

For Question 2,

I don't think that this is necessarily true. For example what if you have map $T: \mathbb{R}^{2}\rightarrow \mathbb{R}^{2}$ such that any vector it takes in it just transforms to 0 vector meaning for $v\in\mathbb{R}^{2}$ that $$T(v)=0$$ well notice that $$T(u+v)=0 \textrm{ and }T(v)+T(u)=0+0=0$$ along with for $c\in\mathbb{R}$ $$T(cv)=0 \textrm{ and }c(Tv)=c(0)=0$$ Thus $T$ is a linear map, but if you take the subspace $U=\{(a,a):a\in\mathbb{R}\}$ (i.e. line $x=y$) well this means for any $u\in U$ that $T(u)=0$ thus the image of U is just the 0 vector (i.e T(U)={0}) thus this linear map "transforms" a line into a point.

  • $\begingroup$ Also the fact that for linear maps that you only really need to know what happens when they are applied to basis vectors, allows us to construct matrices for linear maps. $\endgroup$
    – Kamster
    Aug 27, 2014 at 5:19
  • $\begingroup$ What can we say about the nonzero transformations? As you have taken zero transformation to describe the answer 2? $\endgroup$
    – Srijan
    Aug 27, 2014 at 5:32
  • $\begingroup$ The problem is lets say you are in $R^{3}$ and you have a subspace that is a plane, well this means there there is basis for this plane lets $u$, $v$. Well there is nothing that says $T$ has to be injective so you could have $T(u)=w$ and $T(v)=w$ thus for any vector $z$ in the plane (i.e. in the set $\{z: z=au+bv \textrm{ for }a,b\in\mathbb{R}\}$) that $T(z)=T(au+bv)=(a+b)w$ thus $T$ inadvertently transforms a plane into a line $\endgroup$
    – Kamster
    Aug 27, 2014 at 5:37
  • $\begingroup$ Now an injective linear map would I think necessarily transform a plane into a plane $\endgroup$
    – Kamster
    Aug 27, 2014 at 5:38
  • $\begingroup$ A possible example of transforming plane to line, looking at $\mathbb{R^{3}}$ again, let $T$ such that $T[(a,b,c)]=(a,0,0)$ thus it just spits out x component. Well if you have the plane thats is made by the x and y axis, when you apply $T$ to it you just get x axis (plane to line) $\endgroup$
    – Kamster
    Aug 27, 2014 at 5:42

I am giving you the intuitive meaning of linear transformation based oon coordinate geometry because it is easy to understand and we learned it in higher secondary school. linear transformation A transformation in which the origin of reference frame doesn't change and new cordinate is obtained is linear function of old coordinate i.e. X'=aX+bY and Y'=cX+bY is called linear transformation. In such stated transformation, the straight line remains straight.

You can now replace term old coordinate with standard basis or standard basis vector and new cordinate with cordinate basis or particular vector.


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