Page 196 - A Course in Linear Algebra with Applications

180 Chapter Six: Linear Transformations

space. That this is the case may be seen from a related linear
transformation.
Given an m x n matrix A, define a linear transformation
7\ from F m to F n by the rule Ti(X) = XA. In this case
Im(Xi) is generated by the images of the elements of the stan-
dard basis of F m, that is, by the rows of A. Hence the image
of T\ equals the row space of A.
It is now time to consider what the kernel and image tell
us about a linear transformation.
Theorem 6.3.2
Let T be a linear transformation from a vector space V to a
vector space W. Then
(i) T is infective if and only ifKer(T) is the zero subspace
ofV;
(ii) T is surjective if and only ifIm(T) = W.
Proof
(i) Assume that T is an injective function. If v is a vector in
the kernel of T, then T(v) = 0 W = T(0 V). Therefore v = 0 V
by injectivity, and Ker(T) = Oy- Conversely, suppose that
Ker(T') = Oy- If vi and v 2 are vectors in V with the property
T(vi) = T(v 2 ), then T( V l - v 2 ) = T( V l ) - T(v 2) = 0 W.
Hence the vector vi — v 2 belongs to Ker(T) and vi = v 2 .
(ii) This is true by definition of surjectivity.
For finite-dimensional vector spaces there is a simple for-
mula which links the dimensions of the kernel and image of a
linear transformation.
Theorem 6.3.3
Let T : V —> W be a linear transformation where V and W
are finite-dimensional vector spaces. Then

dim(Ker(T)) + dim(Im(T)) = dim(F).