**Mapping and Functions**: Let *A* and *B* be arbitrary nonempty sets. Suppose to each element in *A* there is assigned a unique element of *B*; the collection *f* of such assignment is called a mapping ( or map) from A into *B*, and is denoted by

*f:A**⟶**B.*

The set *A* is called the domain of the mapping, and *B* is called the target set.

**Linear mapping:** Let *V* and *U* be vector spaces over the same field *K. A* mapping *F:V**⟶**U* is called a linear mapping or linear transformation if it satisfies the following two conditions:

(i) For any vectors *v, w**∊**V, F(v+w) = F(v) + F(w).*

(ii) for any scalar *k* and vector *v**∊**V, F9uv) = kF(v)*

**Example:** Let *F:R ^{3}*

*⟶*

*R*be the “projection” mapping into the xy-plane, that is, F is the mapping defined by

^{3}*F(x, y, z) = (x, y, 0).*We show that

*F*is linear.

Solution: Let *v =(a, b, c)* and *w = (a**′, b′, c′).*

Then *F(v+w) = F(a+a′, b+b′, c+c′)*

*= (a+a′, b+b′, 0)*

* = (a+b+0)+( a′, b′,0)*

* = F(v)+F(w)*

And, for any scalar,

* F(kv)=F(ka, kb, kc) = (ka, kb, 0) = kF(v)*

Thus F is linear mapping.

**Kernel and image of a linear mapping**

Let *F: V**⟶**U *be a linear mapping. The kernel of *F*, written Ker *F*, is the set of elements in *V* that map into the zero vector 0 in *U*; that is,

Ker *F={v**∊**V: F(v) =0}.*

The image of *F*, written *Im F*, is the set of image points in *U*; that is,

*Im F *= {*u**∊**U*:there exist *v**∊**V *for which *F(v)=u}*

**Singular and non-singular linear mapping, Isomorphism:**

Let *F: V**⟶**U *be a linear mapping. Recall that *F(0) = 0*. F is said to be singular if the image of some nonzero vector v is o, that is, if there exist *v**≠0 *such that *F(v) =0. *Thus *F: V**⟶**U *is non singular if the zero vector 0 is the only vector whose image under *F* is 0(zero) or, in other words

*Ker F= {0}.*

**Example:** Consider the projection map *F:R ^{3}*

*⟶*

*R*and the rotation map

^{3}*G:R*

^{3}*⟶*

*R*

^{3}.Let *F:R ^{3}*

*⟶*

*R*be the projection of a vector

^{3}*v*into the xy plane; that is

*F(x, y, 0) = ( x, y, 0)*

Clearly the image of *F* is the entire xy-plane, i.e., points of the form

( x, y, 0). Moreover, the kernel of *F *is the z-axis, i.e., points of the form (0, 0, c). That is,

*Im F= *{(a, b, c):c=0}=*xy-*plane. and *Ker F *= *{(a, b, c):a=0, b=0}= z-*axis.

Since the kernel of *F* is the z-axis, *F* is singular. On the other hand, the kernel of *G* consists only of the zero vector 0. Thus *G* is nonsingular.