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Linear Mapping| Linear Algebra

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:AB.

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:VU is called a linear mapping or linear transformation if it satisfies the following two conditions:

(i) For any vectors v, wV, F(v+w) = F(v) + F(w).

(ii) for any scalar k and vector vV, F9uv) = kF(v)

Example: Let F:R3R3 be the “projection” mapping into the xy-plane, that is, F is the mapping defined by 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: VU 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={vV: F(v) =0}.

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

Im F = {uU:there exist vV for which F(v)=u}

Singular and non-singular linear mapping, Isomorphism:

Let F: VU 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: VU 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:R3 R3 and the rotation map G:R3 R3.

Let F:R3 R3 be the projection of a vector 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.

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