A First Course in Linear Algebra, 2017a

514 Vector Spaces
The values of a,b,c,d that make this true are given by solutions to the system
a − b = 0
c + d = 0
The solution is a = s,b = s,c = t,d = −t where s,t are scalars. We can describe ker(T ) as follows.
{[ ]} {[ ] [ ]}
s s
1 1 0 0
ker(T )=
= span ,
t −t
0 0 1 −1
It is clear that this set is linearly independent and therefore forms a basis for ker(T ).
Wenowwishtofindabasisforim(T).WecanwritetheimageofT as
{[ ]}
a − b
im(T )=
c + d
Notice that this can be written as
{[
1
span
0
] [
−1
,
0
] [
0
,
1
] [
0
,
1
]}
However this is clearly not linearly independent. By removing vectors from the set to create an independent
set gives a basis of im(T). {[ ] [ ]}
1 0
,
0 1
Notice that these vectors have the same span as the set above but are now linearly independent.
♠
A major result is the relation between the dimension of the kernel and dimension of the image of a
linear transformation. A special case was done earlier in the context of matrices. Recall that for an m × n
matrix A, it was the case that the dimension of the kernel of A added to the rank of A equals n.
Theorem 9.81: Dimension of Kernel + Image
Let T : V → W be a linear transformation where V ,W are vector spaces. Suppose the dimension of
V is n. Thenn = dim(ker(T )) + dim(im(T )).
Proof. From Proposition 9.78, im(T ) is a subspace of W. By Theorem 9.48, there exists a basis for
im(T ),{T (⃗v 1 ),···,T (⃗v r )}. Similarly, there is a basis for ker(T ),{⃗u 1 ,···,⃗u s }.Thenif⃗v ∈ V, thereexist
scalars c i such that
T (⃗v)=
r
∑
i=1
c i T (⃗v i )
Hence T (⃗v − ∑ r i=1 c i⃗v i )=0. It follows that⃗v − ∑ r i=1 c i⃗v i is in ker(T ). Hence there are scalars a i such that
⃗v −
r
∑
i=1
c i ⃗v i =
s
∑ a j ⃗u j
j=1