Page 59 - Linear Algebra Done Right

Null Spaces and Ranges
3.4
If V is finite dimensional and T ∈L(V, W), then
Theorem:
range T is a finite-dimensional subspace of W and
dim V = dim null T + dim range T. 45
Proof: Suppose that V is a finite-dimensional vector space and
T ∈L(V, W). Let (u 1 ,...,u m ) be a basis of null T; thus dim null T = m.
The linearly independent list (u 1 ,...,u m ) can be extended to a ba-
sis (u 1 ,...,u m ,w 1 ,...,w n ) of V (by 2.12). Thus dim V = m + n,
and to complete the proof, we need only show that range T is finite
dimensional and dim range T = n. We will do this by proving that
(Tw 1 ,...,Tw n ) is a basis of range T.
Let v ∈ V. Because (u 1 ,...,u m ,w 1 ,...,w n ) spans V, we can write

v = a 1 u 1 +· · ·+ a m u m + b 1 w 1 +· · ·+ b n w n ,

where the a’s and b’s are in F. Applying T to both sides of this equation,
we get
Tv = b 1 Tw 1 + ··· + b n Tw n ,

where the terms of the form Tu j disappeared because each u j ∈ null T.
The last equation implies that (Tw 1 ,...,Tw n ) spans range T. In par-
ticular, range T is finite dimensional.
To show that (Tw 1 ,...,Tw n ) is linearly independent, suppose that
c 1 ,...,c n ∈ F and

c 1 Tw 1 +· · ·+ c n Tw n = 0.

Then
T(c 1 w 1 + ··· + c n w n ) = 0,
and hence
c 1 w 1 +· · ·+ c n w n ∈ null T.
Because (u 1 ,...,u m ) spans null T, we can write

c 1 w 1 +· · ·+ c n w n = d 1 u 1 +· · ·+ d m u m ,

where the d’s are in F. This equation implies that all the c’s (and d’s)
are 0 (because (u 1 ,...,u m ,w 1 ,...,w n ) is linearly independent). Thus
(Tw 1 ,...,Tw n ) is linearly independent and hence is a basis for range T,
as desired.