Properties of Linear Transformations
Theorem: If V and W are vector spaces, T: V  W is a linear transformation, and u and
v are vectors in V, then T has the following properties:
1) T(0) = 0;
2) T(- v) = - T(v);
3) T( u - v) = T(u) - T(v).
Linear transformations preserve linear combinations: Let T: V  W be a linear
transformation, and let v1 , v 2 ,..., v n be vectors in V. If
v  c1 v 1  c 2 v 2  ...  c n v n .
Then,
T ( v)  c1T ( v 1 )  c 2 T ( v 2 )  ...  c nT ( v n )
Thus,
A linear transformation defined on V is completely determined by its values on a set
of basis vectors for V:
Example: Consider the two vectors v 1 = (2, 4) and v 2 = (4, 2)   2 . These vectors are
a basis for  2 . Suppose we want to find the linear transformation T:  2   2 such
that,
T ( v1 )  w1  (3, 5) and T ( v 2 )  w 2  (3,  2)
We know that there is a 2  2 matrix A such that T is defined by T ( v )  Av .
Now
Av1 v 2   Av1 Av 2   T ( v1 ) T ( v 2 )  w1 w 2 
Thus,
2
A
4
4 3
3

.

2 5  2
So,
3  2
3
A

5  2 4
4
2
1
.5

 1.5
.5
2
Hence,
.5  x 
.5
T ( x, y )  
    (.5 x  .5 y, - 1.5x  2y)
- 1.5 2  y 
Definition: The linear transformation T : V  W is one-to-one if no two different
vectors have the same image in W.
Definition: The transformation T : V  W is onto if every vector in W is the image of at
least one vector in V.
Examples:
1. The transformation T :  2   3 defined by T ( x, y )  ( x, y, 1) is one-to-one but is not
onto. Why is transformation not onto?
2. The transformation T :  3  1 defined by T ( x, y, z )  x is onto but is not one-toone. Why is the transformation not one-to-one?
Definition: The linear transformation T : V  W is an isomorphism if it is both one-toone and onto. In this case, given w  W , there exists exactly one vector v  V , such that
T ( v )  w.
Definition: The vector space V is isomorphic to the vector space W if there exists an
isomorphism T : V  W .
Note: If T : V  W is an isomorphism, then T 1 : W  V is defined because T is one-toone and onto. Also, T 1 : W  V is a linear transformation and therefore an
isomorphism. Hence, the vector space W is isomorphic to the vector space V. In such a
case we say that the vector spaces V and W are isomorphic.
Exercise: Show that an n-dimensional vector space is isomorphic to  n (This implies
that any two n-dimensional vector spaces are isomorphic).
Definition: Let T : V  W be a linear transformation. The set of all vectors v in V such
that T(v) = 0 is called the kernel of the linear transformation T, and is denoted by ker(T).
(Note: ker(T) is a subspace of V)
Exercise: Show that the linear transformation T : V  W is one-to-one if and only if
ker(T) ={0}(the zero subspace of V).
Definition: Let T : V  W be a linear transformation. The set of all vectors in W that are
images under T of vectors in V, is called the range of the linear transformation T and is
denoted by range(T).
(Note: range(T) is a subspace of W)
Result: The mapping T : V  W is onto if and only if range(T) = W.
Note: It the range(T) is finite dimensional, then its dimension is called the rank of T,
denoted by rank(T).
Consider the linear transformation T :  n   m given by T ( x)  Ax , where A is m  n ,
and x   n . Then
1. ker (T) = Null(A), i.e., ker (T) is the solution space of the homogeneous linear system
Ax = 0.
2. range(T) = Col(A).
3. rank(T) = dim Col(A) = rank(A).
4. rank(T) + Null(A) = n
Example: The 3 5 matrix A of a linear transformation
8
1  2  2

A  2 3
1
3
1 1
1
4
1
rref (A)  0
0
T :  5   3 is given by
 1
11
 4 
0
0
16/3
1
0
0
1
9/ 4
11 / 12
3
4 
 5
The pivot columns are the first, second, and third.
Hence a basis for Col(A) = {(1, 2, 1), (-2, 3, 1), (-2, -1, 1)}, i.e. rank(T) = 3.
To determine ker(T): Solve
16
u  3v  0
3
9
y  u  4v  0
4
11
z  u  5v  0
12
x
Hence,
x
16
a  3b
3
9
 x   16 / 3 
3 
a  4b
 y  9 / 4 
  4
4




 
11




z   a  5b  z   11 / 12 a  5 b
12
  

 
u  1


0 
ua
v  0

1 
vb
y
Hence a basis for ker(T) = {(-16/3, 9/4, -11/12, 1, 0), (3, -4, 5, 0, 1)} and dim ker(T) = 2.