What is the difference between Linear Mapping and general Mappings

Question

Is there a difference between two terms or can they just be used interchangeably.

If there is a difference, what is it?

I hope this is not too general of a question, its just that one of the tasks in my textbook is asking to prove which mappings are linear. And the word linear is what confuses me, because to prove linearity all I need is this $$L(s\vec{x}+t\vec{y})=sL(\vec{x})+tL(\vec{y})$$

Which looks like all we are saying is that a linear mapping is the one that takes the old one and multiplies it by a scalar. But I'm not sure if I'm understanding this correctly, so need help!

Answer 1

A linear map $T$ over a vector space $V$ over a field $K$ is such that $T(x+y)=T(x)+T(y)$ and $T(\lambda x)=\lambda T(x)$ for every $x,y \in V$ and $\lambda \in K$. Not all mappings are like that.

Answer 2

A linear mapping between two vector spaces $V$ and $W$ can always be represented by a transformation matrix $A$, i.e. $f\colon V\rightarrow W$ with $x\mapsto Ax$ (at least in the finite dimensional case). This is not possible for general mappings. To be a a bit more precise, for a $K$-linear mapping the conditions $f(x+y)=f(x)+f(y)$ and $f(\lambda x)=\lambda f(x)$ need to be satisfied for all $x,y\in V, \lambda\in K.$

Answer 3

No, they are not the same. A linear map is a map $f$ between two $K$-vectorspaces $V$ and $W$, that satisfies additivity ($f(x+y)=f(x)+f(y)$ $\forall x,y \in V$) and homogenity of degree 1 ($f(c*x)=c f(x)$ $\forall x \in V$ and $c \in K$). So trivially, a linear is a map, yes. But not all maps have these properties. A general map doesn't have to be defined between two vectorspaces. General maps can be defined between arbitrary sets.

Answer 4

A "mapping" is just a function. A "linear mapping" is a function with the extra condition that $$L(s \vec x + t \vec y) = s L(\vec x) + t L(\vec y)$$ as you wrote.

Example 1: Consider the mapping $f: \mathbb R \to \mathbb R$ given by $f(x) = 3x$. This is a linear mapping, because \begin{align*} f(ax + by) &= 3 \cdot (ax + by) \\ &= 3ax + 3by \\ &= a \cdot 3x + b \cdot 3y \\ &= a \cdot f(x) + b \cdot f(y). \end{align*}

Example 2: Consider the mapping $f: \mathbb R \to \mathbb R$ given by $f(x) = 3x + 5$. This is not a linear mapping, because \begin{align*} f(2 + 3) &= f(5) = 20 \\ & \neq f(2) + f(3) = 11 + 14 = 25. \end{align*} Notice that here, the "scalar" constants are each $1$. In fact, we could have more quickly checked that this was nonlinear; any linear map must send $\vec 0$ to $\vec 0$, and this function doesn't do that.