1.9 The Matrix of a Linear Transformation
Definition: In Rn, the n x n identity matrix is a
square matrix with 1’s along the main
diagonal, and 0’s elsewhere: In R3,
1 0 0
I 3  0 1 0
0 0 1 . The vectors that make up this
matrix are
1
0 
0
e1  0, e 2  1, e 3  0
0
0
1 a In general, the
i-th column is labeled ei and Inx = x.
The linear transformation T: R3 → Rn is
completely determined by what it does to e1,
e2, and e3.
By the definition of linear, if T: Rm → Rn is a
linear transformation, then
T(0) = 0
And
T(cu + dv) = cT(u) + dT(v)
The generalized result for Rn:
T c1v1  c2 v 2    c p v p  
c1T v1   c2T v 2     c pT v p 
We use this to show that a transformation is
linear.
Example: Let T: R2 → R3 where
2
5 
T e1    3, T e 2   0
 4 
1
If we want to find the matrix A associated to
T, we can compute T(x) for any x by using:
 x1 
x 
 x2 
1 
0 
x1    x2  
0 
1 
x1e1  x2 e 2
so,
T x   T x1e1  x2e 2 
 x1T e1   x2T e 2 
2
5  2 x1  5 x2 
x1  3  x2 0   3x1  0 x2 
 4 
1  4 x1  1x2 
 2 5
 x1 


   3 0  
x2 

 4 1
So
 2 5


A    3 0
 4 1
In general, T x  Ax  T e1  T e2 x
To get A, replace the identity matrix with
T e1  T e2 
Theorem 10: Let
T: Rn → Rm be a linear transformation. Then
there exists a unique matrix A such that
T x   Ax  x  Rn.
In fact, A is the m x n matrix whose j-th
column is the vector T e j  where e j is the j-th
column of the identity matrix In.
A  T e1  T e2   T en 
Example: Given:
 x1  2 x2 
 x1  
T x   A    4 x1 
 x2  3x  2 x 
2
 1
Find A.
Note that T: R2 → R3 and
1  2 
A  T e1  T e 2   4 0 
3 2 
Example:
Find the standard matrix of the linear
transformation
T: R2 → R2 which rotates a point about the

origin through an angle of 4 radians.
Note
 2
 1   2 
T e1   T       
 0    2 
 2 

2
 0     2 
T e 2   T      

 1   2 
 2 
 2
A  T e1  T e 2    2
 2
 2
 2 
2
2 
2 
In general, the matrix of the linear
transformation T: R2 → R2 which rotates a
point about the origin through an angle of 
radians is given by the matrix:
cos   sin  
A

 sin  cos  
Examples: Reflection through the x2-axis
- 1
0

0

1
Reflection through x2 = -x1
 0 - 1
- 1 0 


Vertical contraction by a factor of k (0 < k < 1)
1 0 
0 k 


and expansion (k > 1)
1 0 
0 k 


Vertical sheer
k < 0 and k > 0
 1 0
k 1


Projection onto the x1-axis
1 0
0 0 


Note: this last mapping takes any point in R2
and moves it down to the x1-axis, so the
range is not all of R2. We distinguish between
transformations whose range is all of the
codomain and whose range is not.
Definition: A mapping T: Rn → Rm is said to
be onto if each b in Rm is the image of at least
one x in Rn, that is T(x) = b has at least one
solution.
Definition: T: Rn → Rm is said to be one to one
if each b in Rm is the image of at most one x
in Rn.
Example: T: R2 → R2 the expansion (k > 1)
1 0 
0 k  is one-to-one and onto since every


 x1 
 
vector  x2  in R2 is the image of the vector
 x1 
1 
 x2  .
k 
Example: T: Rn → Rm defined by T(x) = 0x is
neither one-to-one nor onto since it maps all
of Rn to the single vector 0 in Rm.
Example: Let T: R4 → R3 defined by T(x) = Ax
where
1  1 3 7 
A  0 2 5 4 
0 0 3  6
Is T onto? In other words: Does the equation
Ax = b have a solution for all b in R3?
By Th. 1.4, this equation is consistent iff A
has a pivot in each row. The matrix is in row
echelon form, so we can see that there is a
pivot in each row, thus the equation Ax = b is
consistent, so T is onto.
Is T one-to-one? In other words: Is the
solution Ax = b unique for all b in R3?
The row echelon form of the matrix shows
that x4 is free, so by Th. 1.2, there are
infinitely many solutions. Therefore, the
solution is not unique, so T is not one-to-one.
The next theorem follows from this.
Theorem 1.11: Let
T: Rn → Rm be a linear transformation. Then
T is one-to-one iff the equation Ax = 0 has
only the trivial solution.
Proof:
T is 1 to1 so T(x) = 0 has at most one
solution. Since T is linear, T(0) = 0.
Thus, T(x) = 0 implies that x = 0.
So, T(x) = 0 has only the trivial solution.
Want to prove that if T(x) = 0 has only the
trivial solution, T(x) is one to one.
We will prove the contrapositive: If T is not 1
to 1, T(x) = 0 has more than the trivial solution
Assume T is not 1 to 1.
Then  b in Rn such that b is the image of two
different vectors, u and v.
That is,
T(u) = b and T(v) = b
Since T is linear,
T(u – v) = T(u) – T(v)
= b – b = 0.
But the vector u – v is not 0 since u ≠ v, so
we have two vectors whose image under T is
0. Therefore Ax = 0 has more than the trivial
solution.
So, either the two conditions in the theorem
are both true or they are both false. This
proves the other direction of the “if and only if”
and we can say that T is one-to-one iff the
equation Ax = 0 has only the trivial solution.
Theorem 1.12: Let
T: Rn → Rm be a linear transformation and A
be the standard matrix for T. Then
a) T maps Rn onto Rm if and only if the
columns of A span Rm.
b) T is one-to-one if and only if the
columns of A are linearly independent.
Proof: (a)
By the definition of onto, T maps Rn onto Rm
if and only if
the equation Ax = b has at least one solution
for all b Rm
if and only if
the columns of A span Rm By Th. 4,.
So, T maps Rn onto Rm if and only if the
columns of A span Rm.
b) By Th 11, T is one-to-one
if and only if
the equation Ax = 0 or T(x) = 0 has only the
trivial solution.
And Ax = 0 has only the trivial solution
if and only if the columns of A are linearly
independent by the definition of linear
independence: x1a1  x2a2    xnan  0 has
only the trivial solution.
T is one-to-one if and only if the columns of A
are linearly independent.
Example: (Cheesy notation, but you may see
it in the text.)
Let T x1, x2   2 x1  x2 , x1,4 x1  3x2 , x2 
Is T one-to-one and onto?
First, we write this with better notation to help
us find the matrix A that defines T.
 2 x1  x2 
  x1    x1 
T     
  x2   4 x1  3x2 


x
2


Factoring out x1 and x2 gives:
2
1

4

0
 1
0   x1 
3   x2 

1
A has only 2 columns and they aren’t scalar
multiples of each other, so they are linearly
independent, thus by Th. 12b, T is 1 to 1.
T is onto iff the columns of A span R4.
By Th 4, the columns of A span R4 only if
there are 4 rows with pivot positions. This is
impossible since the A has only two columns,
so T cannot be onto.