Page 197 - A Course in Linear Algebra with Applications

6.3: Kernel, Image and Isomorphism 181

Proof
Here we may assume that V is not the zero space; otherwise
the statement is true for obvious reasons. By 5.1.4 it is possi-
ble to choose a basis v i , . . . , v n of V such that part of it is a
basis of Ker(T), say v i , . . . , v r ; here of course

n = dim(V) > r = dim(Ker(T)).

We claim that the vectors T(v T . + 1 ),..., T(v n) are linearly in-
dependent. For if c r + iT(v r + i) + • • • + c nT(v n) = Ow for
some scalars Ci, then T(c r+iv r+i + • • • + c n v n ) = Oiy, so
that c r+iv r_|_i + • • • + c n v n belongs to Ker(T) and is there-
fore expressible as a linear combination of v i , . . . , v r . But
v i , . . . , v r , . . . , v n are certainly linearly independent. Hence
c r + i,..., c n are all zero and our claim is established.
On the other hand, the vectors T(v r + 1 ),... ,T(v n ) by
themselves generate Im(T) since T(vi) = • • • = T(v r ) = Ow',
hence T(v r + i),... T(v n ) form a basis of Im(T). It follows
,
that
dim(Im(T)) = n-r = dim(F) - dim(Ker(T)),

from which the formula follows.
The dimension formula is in fact a generalization of some-
thing that we already know. For suppose we apply the for-
mula to the linear transformation T : F n —• F m defined by
T(X) = AX, where A is an m x n matrix. Making the inter-
pretations of Ker(T) and Im(T) as the null space and column
space of A, we deduce that the sum of the dimensions of the
null space and column space of A equals n. This is essentially
the content of 5.1.7 and 5.2.4.
Isomorphism
Because of 6.3.2 we can tell whether a linear transforma-
tion T : V —•> W is bijective. And in view of 6.1.3 this is the
same as asking whether T has an inverse. A bijective linear
transformation is called an isomorphism.