A First Course in Linear Algebra, 2017a

5.7. The Kernel And Image Of A Linear Map 307
Recall that a linear transformation T is called one to one if and only if T (⃗x)=⃗0 implies⃗x =⃗0. Using
the concept of kernel, we can state this theorem in another way.
Theorem 5.51: One to One and Kernel
Let T be a linear transformation where ker(T ) is the kernel of T .ThenT is one to one if and only
if ker(T ) consists of only the zero vector.
♠
A major result is the relation between the dimension of the kernel and dimension of the image of a
linear transformation. In the previous example ker(T ) had dimension 2, and im(T ) also had dimension of
2. Consider the following theorem.
Theorem 5.52: Dimension of Kernel and Image
Let T : V → W be a linear transformation where V ,W are subspaces of R n . Suppose the dimension
of V is m. Then
m = dim(ker(T )) + dim(im(T ))
Proof. From Proposition 5.49, im(T ) is a subspace of W. We know that there exists a basis for im(T ),
written {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
Hence⃗v = ∑ r i=1 c i⃗v i + ∑ s j=1 a j⃗u j .Since⃗v is arbitrary, it follows that
V = span{⃗u 1 ,···,⃗u s ,⃗v 1 ,···,⃗v r }
If the vectors {⃗u 1 ,···,⃗u s ,⃗v 1 ,···,⃗v r } are linearly independent, then it will follow that this set is a basis.
Suppose then that
r s
∑ c i ⃗v i + ∑ a j ⃗u j = 0
i=1 j=1
Apply T to both sides to obtain
r
∑
i=1
c i T (⃗v i )+
s
∑ a j T (⃗u) j =
j=1
r
∑
i=1
c i T (⃗v i )=0