LINEAR ALGEBRA
W W L CHEN
c
W W L Chen, 1997, 2008.
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Chapter 8
LINEAR TRANSFORMATIONS
8.1. Euclidean Linear Transformations
By a transformation from Rn into Rm , we mean a function of the type T : Rn → Rm , with domain Rn
and codomain Rm . For every vector x ∈ Rn , the vector T (x) ∈ Rm is called the image of x under the
transformation T , and the set
R(T ) = {T (x) : x ∈ Rn },
of all images under T , is called the range of the transformation T .
Remark. For our convenience later, we have chosen to use R(T ) instead of the usual T (Rn ) to denote
the range of the transformation T .
For every x = (x1 , . . . , xn ) ∈ Rn , we can write
T (x) = T (x1 , . . . , xn ) = (y1 , . . . , ym ).
Here, for every i = 1, . . . , m, we have
yi = Ti (x1 , . . . , xn ),
(1)
where Ti : Rn → R is a real valued function.
Definition. A transformation T : Rn → Rm is called a linear transformation if there exists a real
matrix


a11 . . . a1n
.
.
.. 
A =  ..
am1 . . . amn
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W W L Chen, 1997, 2008
such that for every x = (x1 , . . . , xn ) ∈ Rn , we have T (x1 , . . . , xn ) = (y1 , . . . , ym ), where
y1 = a11 x1 + . . . + a1n xn ,
..
.
ym = am1 x1 + . . . + amn xn ,
or, in matrix notation,

 
y1
a11
 ..   ..
=
 . 
.
am1
ym
 
a1n
x1
..   .. 
.
.
.
...
...
amn
(2)
xn
The matrix A is called the standard matrix for the linear transformation T .
Remarks. (1) In other words, a transformation T : Rn → Rm is linear if the equation (1) for every
i = 1, . . . , m is linear.
(2) If we write x ∈ Rn and y ∈ Rm as column matrices, then (2) can be written in the form y = Ax,
and so the linear transformation T can be interpreted as multiplication of x ∈ Rn by the standard
matrix A.
Definition. A linear transformation T : Rn → Rm is said to be a linear operator if n = m. In this case,
we say that T is a linear operator on Rn .
Example 8.1.1. The linear transformation T : R5 → R3 , defined by the equations
y1 = 2x1 + 3x2 + 5x3 + 7x4 − 9x5 ,
y2 =
3x2 + 4x3
y3 = x1
+ 2x5 ,
+ 3x3 − 2x4
,
can be expressed in matrix form as



y1
2
 y2  =  0
1
y3
3
3
0
5
4
3
7
0
−2


x1
−9  x2 
 
2   x3  .
 
0
x4
x5

If (x1 , x2 , x3 , x4 , x5 ) = (1, 0, 1, 0, 1), then



y1
2 3 5
 y2  =  0 3 4
y3
1 0 3
7
0
−2
 
1


−9  0 
−2
 
2 1 =  6 ,
 
0
0
4
1

so that T (1, 0, 1, 0, 1) = (−2, 6, 4).
Example 8.1.2. Suppose that A is the zero m × n matrix. The linear transformation T : Rn → Rm ,
where T (x) = Ax for every x ∈ Rn , is the zero transformation from Rn into Rm . Clearly T (x) = 0 for
every x ∈ Rn .
Example 8.1.3. Suppose that I is the identity n × n matrix. The linear operator T : Rn → Rn , where
T (x) = Ix for every x ∈ Rn , is the identity operator on Rn . Clearly T (x) = x for every x ∈ Rn .
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PROPOSITION 8A. Suppose that T : Rn → Rm is a linear transformation, and that {e1 , . . . , en } is
the standard basis for Rn . Then the standard matrix for T is given by
A = ( T (e1 )
...
T (en ) ) ,
where T (ej ) is a column matrix for every j = 1, . . . , n.
Proof. This follows immediately from (2). 8.2. Linear Operators on R2
In this section, we consider the special case when n = m = 2, and study linear operators on R2 . For
every x ∈ R2 , we shall write x = (x1 , x2 ).
Example 8.2.1. Consider reflection across the x2 -axis, so that T (x1 , x2 ) = (−x1 , x2 ). Clearly we have
−1
0
T (e1 ) =
and
T (e2 ) =
,
0
1
and so it follows from Proposition 8A that the standard matrix is given by
−1 0
A=
.
0 1
It is not difficult to see that the standard matrices for reflection across the x1 -axis and across the line
x1 = x2 are given respectively by
1 0
0 1
A=
and
A=
.
0 −1
1 0
Also, the standard matrix for reflection across the origin is given by
−1 0
A=
.
0 −1
We give a summary in the table below:
Linear operator
Reflection across x2 -axis
Reflection across x1 -axis
Reflection across x1 = x2
Reflection across origin
Equations
n y = −x
1
y2 = x2
1
ny = x
1
1
y2 = −x2
ny = x
1
2
y2 = x1
n y = −x
1
1
y2 = −x2
Standard matrix
−1 0
0 1
1 0
0 −1
0 1
1 0
−1 0
0 −1
Example 8.2.2. For orthogonal projection onto the x1 -axis, we have T (x1 , x2 ) = (x1 , 0), with standard
matrix
1 0
A=
.
0 0
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Similarly, the standard matrix for orthogonal projection onto the x2 -axis is given by
0 0
A=
.
0 1
We give a summary in the table below:
Linear operator
Orthogonal projection onto x1 -axis
Equations
ny = x
1
Orthogonal projection onto x2 -axis
y2 = 0
Standard matrix
1 0
0 0
0 0
0 1
1
y1 = 0
y2 = x2
Example 8.2.3. For anticlockwise rotation by an angle θ, we have T (x1 , x2 ) = (y1 , y2 ), where
y1 + iy2 = (x1 + ix2 )(cos θ + i sin θ),
and so
y1
y2
=
cos θ
sin θ
It follows that the standard matrix is given by
cos θ
A=
sin θ
− sin θ
cos θ
− sin θ
cos θ
x1
x2
.
.
We give a summary in the table below:
Linear operator
Anticlockwise rotation by angle θ
Equations
y1 = x1 cos θ − x2 sin θ
y2 = x1 sin θ + x2 cos θ
Standard matrix
cos θ − sin θ
sin θ
cos θ
Example 8.2.4. For contraction or dilation by a non-negative scalar k, we have T (x1 , x2 ) = (kx1 , kx2 ),
with standard matrix
k 0
A=
.
0 k
The operator is called a contraction if 0 < k < 1 and a dilation if k > 1, and can be extended to negative
values of k by noting that for k < 0, we have
k 0
−1 0
−k 0
=
.
0 k
0 −1
0 −k
This describes contraction or dilation by non-negative scalar −k followed by reflection across the origin.
We give a summary in the table below:
Linear operator
Contraction or dilation by factor k
Chapter 8 : Linear Transformations
Equations
y1 = kx1
y2 = kx2
Standard matrix
k 0
0 k
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Linear
Linear Algebra
Algebra
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2008
WW
WL
L Chen,
Chen, 1997,
1997, 2006
2006
Example 8.2.5. For expansion or compression in the x11 -direction by a positive factor k, we have
T (x11 , x22 ) = (kx11 , x22 ), with standard matrix
"
!
k 0
A=
.
0 1
This can be extended to negative values of k by noting that for
!
"
"
!
!
k
−k
k 00 = −1
−1 00
−k
=
00 11
00 11
00
This
This describes
describes expansion
expansion or
or compression
compression in
in the
the x
x11 -direction
-direction
across
the
x
-axis.
Similarly,
for
expansion
or
compression
2
across the x2 -axis. Similarly, for expansion or compression
we
we have
have the
the standard
standard matrix
matrix
!
"
11 00
A
A=
= 0 k ..
0 k
k < 0, we have
"
00
.
11 .
by
by positive
positive factor
factor −k
−k followed
followed by
by reflection
reflection
in
the
x
-direction
by
a
non-zero
2
in the x2 -direction by a non-zero factor
factor k,
k,
We
We give
give aa summary
summary in
in the
the table
table below:
below:
Linear
Linear operator
operator
Equations
Equations
#
y111 = kx111
yy222 =
=x
x222
$
n yy11 =
x11
1 = x1
yy222 =
= kx
kx222
Expansion
Expansion or
or compression
compression in
in x
x111 -direction
-direction
Expansion
Expansion or
or compression
compression in
in x
x222 -direction
-direction
Standard
Standard matrix
matrix
!
"
k 0
00 11
!
1 0"
1 0
00 k
k
Example
Example 8.2.6.
8.2.6. For
For shears
shears in
in the
the x
x111 -direction
-direction with
with factor
factor k,
k, we
we have
have T
T (x
(x111 ,, x
x222 )) =
= (x
(x111 +
+ kx
kx222 ,, x
x222 ),
), with
with
standard
matrix
standard matrix
!
1 k"
1 k
A
=
A = 0 1 ..
0 1
For
For the
the case
case k
k=
= 1,
1, we
we have
have the
the following.
following.
•
•
•
•
T
T
(k=1)
(k=1)
•
•
•
•
For
For the
the case
case k =
= −1,
−1, we have
have the
the following.
following.
•
•
•
•
Chapter
Chapter 8
8 :: Linear
Linear Transformations
Transformations
T
T
(k=−1)
(k=−1)
•
•
•
•
page
page 5
5 of
of 35
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Similarly, for shears in the x2 -direction with factor k, we have standard matrix
A=
1
k
0
1
.
We give a summary in the table below:
Linear operator
Shear in x1 -direction
Shear in x2 -direction
Equations
y1 = x1 + kx2
y2 = x2
ny = x
1
1
y2 = kx1 + x2
Standard matrix
1 k
0 1
1 0
k 1
Example 8.2.7. Consider a linear operator T : R2 → R2 which consists of a reflection across the x2 -axis,
followed by a shear in the x1 -direction with factor 3 and then reflection across the x1 -axis. To find the
standard matrix, consider the effect of T on a standard basis {e1 , e2 } of R2 . Note that
1
−1
−1
−1
7→
7→
7→
= T (e1 ),
0
0
0
0
0
0
3
3
e2 =
7→
7→
7→
= T (e2 ),
1
1
1
−1
e1 =
so it follows from Proposition 8A that the standard matrix for T is
A=
−1
0
3
−1
.
Let us summarize the above and consider a few special cases. We have the following table of invertible
linear operators with k 6= 0. Clearly, if A is the standard matrix for an invertible linear operator T , then
the inverse matrix A−1 is the standard matrix for the inverse linear operator T −1 .
Linear operator T
Reflection across
line x1 =x2
Expansion or compression
in x1 −direction
Expansion or compression
in x2 −direction
Shear
in x1 −direction
Shear
in x2 −direction
Standard
0
1
k
0
1
0
1
0
1
k
matrix A
1
0
0
1
0
k
k
1
0
1
Inverse matrix A−1
0 1
1 0
−1
k
0
0
1
1
0
0 k −1
1 −k
0 1
1 0
−k 1
Linear operator T −1
Reflection across
line x1 =x2
Expansion or compression
in x1 −direction
Expansion or compression
in x2 −direction
Shear
in x1 −direction
Shear
in x2 −direction
Next, let us consider the question of elementary row operations on 2 × 2 matrices. It is not difficult
to see that an elementary row operation performed on a 2 × 2 matrix A has the effect of multiplying the
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matrix A by some elementary matrix E to give the product EA. We have the following table.
Elementary row operation
Elementary matrix E
0 1
1 0
k 0
0 1
1 0
0 k
1 k
0 1
1 0
k 1
Interchanging the two rows
Multiplying row 1 by non-zero factor k
Multiplying row 2 by non-zero factor k
Adding k times row 2 to row 1
Adding k times row 1 to row 2
Now, we know that any invertible matrix A can be reduced to the identity matrix by a finite number of
elementary row operations. In other words, there exist a finite number of elementary matrices E1 , . . . , Es
of the types above with various non-zero values of k such that
Es . . . E1 A = I,
so that
A = E1−1 . . . Es−1 .
We have proved the following result.
PROPOSITION 8B. Suppose that the linear operator T : R2 → R2 has standard matrix A, where A is
invertible. Then T is the product of a succession of finitely many reflections, expansions, compressions
and shears.
In fact, we can prove the following result concerning images of straight lines.
PROPOSITION 8C. Suppose that the linear operator T : R2 → R2 has standard matrix A, where A
is invertible. Then
(a) the image under T of a straight line is a straight line;
(b) the image under T of a straight line through the origin is a straight line through the origin; and
(c) the images under T of parallel straight lines are parallel straight lines.
Proof. Suppose that T (x1 , x2 ) = (y1 , y2 ). Since A is invertible, we have x = A−1 y, where
x=
x1
x2
and
y=
y1
y2
.
The equation of a straight line is given by αx1 + βx2 = γ or, in matrix form, by
(α
β)
x1
x2
= (γ ).
Hence
(α
Chapter 8 : Linear Transformations
β )A
−1
y1
y2
= (γ ).
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Let
( α0
β0 ) = ( α
Then
( α0
β0 )
y1
y2
β ) A−1 .
= (γ ).
In other words, the image under T of the straight line αx1 + βx2 = γ is α0 y1 + β 0 y2 = γ, clearly another
straight line. This proves (a). To prove (b), note that straight lines through the origin correspond to
γ = 0. To prove (c), note that parallel straight lines correspond to different values of γ for the same
values of α and β. 8.3. Elementary Properties of Euclidean Linear Transformations
In this section, we establish a number of simple properties of euclidean linear transformations.
PROPOSITION 8D. Suppose that T1 : Rn → Rm and T2 : Rm → Rk are linear transformations.
Then T = T2 ◦ T1 : Rn → Rk is also a linear transformation.
Proof. Since T1 and T2 are linear transformations, they have standard matrices A1 and A2 respectively.
In other words, we have T1 (x) = A1 x for every x ∈ Rn and T2 (y) = A2 y for every y ∈ Rm . It follows
that T (x) = T2 (T1 (x)) = A2 A1 x for every x ∈ Rn , so that T has standard matrix A2 A1 . Example 8.3.1. Suppose that T1 : R2 → R2 is anticlockwise rotation by π/2 and T2 : R2 → R2 is
orthogonal projection onto the x1 -axis. Then the respective standard matrices are
0 −1
1 0
A1 =
and
A2 =
.
1 0
0 0
It follows that the standard matrices for T2 ◦ T1 and T1 ◦ T2 are respectively
0 −1
0 0
A2 A1 =
and
A1 A2 =
.
0 0
1 0
Hence T2 ◦ T1 and T1 ◦ T2 are not equal.
Example 8.3.2. Suppose that T1 : R2 → R2 is anticlockwise rotation by θ and T2 : R2 → R2 is
anticlockwise rotation by φ. Then the respective standard matrices are
cos θ − sin θ
cos φ − sin φ
A1 =
and
A2 =
.
sin θ
cos θ
sin φ cos φ
It follows that the standard matrix for T2 ◦ T1 is
cos φ cos θ − sin φ sin θ − cos φ sin θ − sin φ cos θ
cos(φ + θ) − sin(φ + θ)
A2 A1 =
=
.
sin φ cos θ + cos φ sin θ
cos φ cos θ − sin φ sin θ
sin(φ + θ) cos(φ + θ)
Hence T2 ◦ T1 is anticlockwise rotation by φ + θ.
Example 8.3.3. The reader should check that in R2 , reflection across the x1 -axis followed by reflection
across the x2 -axis gives reflection across the origin.
Linear transformations that map distinct vectors to distinct vectors are of special importance.
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Definition. A linear transformation T : Rn → Rm is said to be one-to-one if for every x0 , x00 ∈ Rn , we
have x0 = x00 whenever T (x0 ) = T (x00 ).
Example 8.3.4. If we consider linear operators T : R2 → R2 , then T is one-to-one precisely when the
standard matrix A is invertible. To see this, suppose first of all that A is invertible. If T (x0 ) = T (x00 ),
then Ax0 = Ax00 . Multiplying on the left by A−1 , we obtain x0 = x00 . Suppose next that A is not
invertible. Then there exists x ∈ R2 such that x 6= 0 and Ax = 0. On the other hand, we clearly have
A0 = 0. It follows that T (x) = T (0), so that T is not one-to-one.
PROPOSITION 8E. Suppose that the linear operator T : Rn → Rn has standard matrix A. Then the
following statements are equivalent:
(a) The matrix A is invertible.
(b) The linear operator T is one-to-one.
(c) The range of T is Rn ; in other words, R(T ) = Rn .
Proof. ((a)⇒(b)) Suppose that T (x0 ) = T (x00 ). Then Ax0 = Ax00 . Multiplying on the left by A−1 gives
x0 = x00 .
((b)⇒(a)) Suppose that T is one-to-one. Then the system Ax = 0 has unique solution x = 0 in Rn .
It follows that A can be reduced by elementary row operations to the identity matrix I, and is therefore
invertible.
((a)⇒(c)) For any y ∈ Rn , clearly x = A−1 y satisfies Ax = y, so that T (x) = y.
((c)⇒(a)) Suppose that {e1 , . . . , en } is the standard basis for Rn . Let x1 , . . . , xn ∈ Rn be chosen to
satisfy T (xj ) = ej , so that Axj = ej , for every j = 1, . . . , n. Write
C = ( x1
...
xn ) .
Then AC = I, so that A is invertible. Definition. Suppose that the linear operator T : Rn → Rn has standard matrix A, where A is invertible.
Then the linear operator T −1 : Rn → Rn , defined by T −1 (x) = A−1 x for every x ∈ Rn , is called the
inverse of the linear operator T .
Remark. Clearly T −1 (T (x)) = x and T (T −1 (x)) = x for every x ∈ Rn .
Example 8.3.5. Consider the linear operator T : R2 → R2 , defined by T (x) = Ax for every x ∈ R2 ,
where
1 1
A=
.
1 2
Clearly A is invertible, and
A−1 =
2
−1
−1
1
.
Hence the inverse linear operator is T −1 : R2 → R2 , defined by T −1 (x) = A−1 x for every x ∈ R2 .
Example 8.3.6. Suppose that T : R2 → R2 is anticlockwise rotation by angle θ. The reader should
check that T −1 : R2 → R2 is anticlockwise rotation by angle 2π − θ.
Next, we study the linearity properties of euclidean linear transformations which we shall use later to
discuss linear transformations in arbitrary real vector spaces.
Chapter 8 : Linear Transformations
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PROPOSITION 8F. A transformation T : Rn → Rm is linear if and only if the following two
conditions are satisfied:
(a) For every u, v ∈ Rn , we have T (u + v) = T (u) + T (v).
(b) For every u ∈ Rn and c ∈ R, we have T (cu) = cT (u).
Proof. Suppose first of all that T : Rn → Rm is a linear transformation. Let A be the standard matrix
for T . Then for every u, v ∈ Rn and c ∈ R, we have
T (u + v) = A(u + v) = Au + Av = T (u) + T (v)
and
T (cu) = A(cu) = c(Au) = cT (u).
Suppose now that (a) and (b) hold. To show that T is linear, we need to find a matrix A such that
T (x) = Ax for every x ∈ Rn . Suppose that {e1 , . . . , en } is the standard basis for Rn . As suggested by
Proposition 8A, we write
A = ( T (e1 ) . . .
T (en ) ) ,
where T (ej ) is a column matrix for every j = 1, . . . , n. For any vector


x1
.
x =  .. 
xn
in Rn , we have

Ax = ( T (e1 ) . . .

x1
.
T (en ) )  ..  = x1 T (e1 ) + . . . + xn T (en ).
xn
Using (b) on each summand and then using (a) inductively, we obtain
Ax = T (x1 e1 ) + . . . + T (xn en ) = T (x1 e1 + . . . + xn en ) = T (x)
as required. To conclude our study of euclidean linear transformations, we briefly mention the problem of eigenvalues and eigenvectors of euclidean linear operators.
Definition. Suppose that T : Rn → Rn is a linear operator. Then any real number λ ∈ R is called
an eigenvalue of T if there exists a non-zero vector x ∈ Rn such that T (x) = λx. This non-zero vector
x ∈ Rn is called an eigenvector of T corresponding to the eigenvalue λ.
Remark. Note that the equation T (x) = λx is equivalent to the equation Ax = λx. It follows that
there is no distinction between eigenvalues and eigenvectors of T and those of the standard matrix A.
We therefore do not need to discuss this problem any further.
8.4. General Linear Transformations
Suppose that V and W are real vector spaces. To define a linear transformation from V into W , we are
motivated by Proposition 8F which describes the linearity properties of euclidean linear transformations.
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By a transformation from V into W , we mean a function of the type T : V → W , with domain V
and codomain W . For every vector u ∈ V , the vector T (u) ∈ W is called the image of u under the
transformation T .
Definition. A transformation T : V → W from a real vector space V into a real vector space W is
called a linear transformation if the following two conditions are satisfied:
(LT1) For every u, v ∈ V , we have T (u + v) = T (u) + T (v).
(LT2) For every u ∈ V and c ∈ R, we have T (cu) = cT (u).
Definition. A linear transformation T : V → V from a real vector space V into itself is called a linear
operator on V .
Example 8.4.1. Suppose that V and W are two real vector spaces. The transformation T : V → W ,
where T (u) = 0 for every u ∈ V , is clearly linear, and is called the zero transformation from V to W .
Example 8.4.2. Suppose that V is a real vector space. The transformation I : V → V , where I(u) = u
for every u ∈ V , is clearly linear, and is called the identity operator on V .
Example 8.4.3. Suppose that V is a real vector space, and that k ∈ R is fixed. The transformation
T : V → V , where T (u) = ku for every u ∈ V , is clearly linear. This operator is called a dilation if
k > 1 and a contraction if 0 < k < 1.
Example 8.4.4. Suppose that V is a finite dimensional vector space, with basis {w1 , . . . , wn }. Define a
transformation T : V → Rn as follows. For every u ∈ V , there exists a unique vector (β1 , . . . , βn ) ∈ Rn
such that u = β1 w1 + . . . + βn wn . We let T (u) = (β1 , . . . , βn ). In other words, the transformation T
gives the coordinates of any vector u ∈ V with respect to the given basis {w1 , . . . , wn }. Suppose now
that v = γ1 w1 + . . . + γn wn is another vector in V . Then u + v = (β1 + γ1 )w1 + . . . + (βn + γn )wn , so
that
T (u + v) = (β1 + γ1 , . . . , βn + γn ) = (β1 , . . . , βn ) + (γ1 , . . . , γn ) = T (u) + T (v).
Also, if c ∈ R, then cu = cβ1 w1 + . . . + cβn wn , so that
T (cu) = (cβ1 , . . . , cβn ) = c(β1 , . . . , βn ) = cT (u).
Hence T is a linear transformation. We shall return to this in greater detail in the next section.
Example 8.4.5. Suppose that Pn denotes the vector space of all polynomials with real coefficients and
degree at most n. Define a transformation T : Pn → Pn as follows. For every polynomial
p = p0 + p1 x + . . . + pn xn
in Pn , we let
T (p) = pn + pn−1 x + . . . + p0 xn .
Suppose now that q = q0 + q1 x + . . . + qn xn is another polynomial in Pn . Then
p + q = (p0 + q0 ) + (p1 + q1 )x + . . . + (pn + qn )xn ,
so that
T (p + q) = (pn + qn ) + (pn−1 + qn−1 )x + . . . + (p0 + q0 )xn
= (pn + pn−1 x + . . . + p0 xn ) + (qn + qn−1 x + . . . + q0 xn ) = T (p) + T (q).
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Also, for any c ∈ R, we have cp = cp0 + cp1 x + . . . + cpn xn , so that
T (cp) = cpn + cpn−1 x + . . . + cp0 xn = c(pn + pn−1 x + . . . + p0 xn ) = cT (p).
Hence T is a linear transformation.
Example 8.4.6. Let V denote the vector space of all real valued functions differentiable everywhere in R,
and let W denote the vector space of all real valued functions defined on R. Consider the transformation
T : V → W , where T (f ) = f 0 for every f ∈ V . It is easy to check from properties of derivatives that T
is a linear transformation.
Example 8.4.7. Let V denote the vector space of all real valued functions that are Riemann integrable
over the interval [0, 1]. Consider the transformation T : V → R, where
Z 1
T (f ) =
f (x) dx
0
for every f ∈ V . It is easy to check from properties of the Riemann integral that T is a linear transformation.
Consider a linear transformation T : V → W from a finite dimensional real vector space V into a real
vector space W . Suppose that {v1 , . . . , vn } is a basis of V . Then every u ∈ V can be written uniquely
in the form u = β1 v1 + . . . + βn vn , where β1 , . . . , βn ∈ R. It follows that
T (u) = T (β1 v1 + . . . + βn vn ) = T (β1 v1 ) + . . . + T (βn vn ) = β1 T (v1 ) + . . . + βn T (vn ).
We have therefore proved the following generalization of Proposition 8A.
PROPOSITION 8G. Suppose that T : V → W is a linear transformation from a finite dimensional
real vector space V into a real vector space W . Suppose further that {v1 , . . . , vn } is a basis of V . Then
T is completely determined by T (v1 ), . . . , T (vn ).
Example 8.4.8. Consider a linear transformation T : P2 → R, where T (1) = 1, T (x) = 2 and T (x2 ) = 3.
Since {1, x, x2 } is a basis of P2 , this linear transformation is completely determined. In particular, we
have, for example,
T (5 − 3x + 2x2 ) = 5T (1) − 3T (x) + 2T (x2 ) = 5.
Example 8.4.9. Consider a linear transformation T : R4 → R, where T (1, 0, 0, 0) = 1, T (1, 1, 0, 0) = 2,
T (1, 1, 1, 0) = 3 and T (1, 1, 1, 1) = 4. Since {(1, 0, 0, 0), (1, 1, 0, 0), (1, 1, 1, 0), (1, 1, 1, 1)} is a basis of R4 ,
this linear transformation is completely determined. In particular, we have, for example,
T (6, 4, 3, 1) = T (2(1, 0, 0, 0) + (1, 1, 0, 0) + 2(1, 1, 1, 0) + (1, 1, 1, 1))
= 2T (1, 0, 0, 0) + T (1, 1, 0, 0) + 2T (1, 1, 1, 0) + T (1, 1, 1, 1) = 14.
We also have the following generalization of Proposition 8D.
PROPOSITION 8H. Suppose that V, W, U are real vector spaces. Suppose further that T1 : V → W
and T2 : W → U are linear transformations. Then T = T2 ◦ T1 : V → U is also a linear transformation.
Proof. Suppose that u, v ∈ V . Then
T (u + v) = T2 (T1 (u + v)) = T2 (T1 (u) + T1 (v)) = T2 (T1 (u)) + T2 (T1 (v)) = T (u) + T (v).
Also, if c ∈ R, then
T (cu) = T2 (T1 (cu)) = T2 (cT1 (u)) = cT2 (T1 (u)) = cT (u).
Hence T is a linear transformation. Chapter 8 : Linear Transformations
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8.5. Change of Basis
Suppose that V is a real vector space, with basis B = {u1 , . . . , un }. Then every vector u ∈ V can be
written uniquely as a linear combination
u = β1 u1 + . . . + βn un ,
where β1 , . . . , βn ∈ R.
(3)
It follows that the vector u can be identified with the vector (β1 , . . . , βn ) ∈ Rn .
Definition. Suppose that u ∈ V and (3) holds. Then the matrix
 
β1
 .. 
[u]B =  . 
βn
is called the coordinate matrix of u relative to the basis B = {u1 , . . . , un }.
Example 8.5.1. The vectors
u1 = (1, 2, 1, 0),
u2 = (3, 3, 3, 0),
u3 = (2, −10, 0, 0),
u4 = (−2, 1, −6, 2)
are linearly independent in R4 , and so B = {u1 , u2 , u3 , u4 } is a basis of R4 . It follows that for any
u = (x, y, z, w) ∈ R4 , we can write
u = β1 u1 + β2 u2 + β3 u3 + β4 u4 .
In matrix notation, this becomes
 
1 3
x
 y  2 3
 =
1 3
z
0 0
w

 
β1
2
−2
−10 1   β2 
 ,
0
−6
β3
0
2
β4
so that

 
β1
1
 β2   2
[u]B =   = 
β3
1
β4
0
3
3
3
0
−1  
2
−2
x
−10 1   y 
  .
0
−6
z
0
2
w
Remark. Consider a function φ : V → Rn , where φ(u) = [u]B for every u ∈ V . It is not difficult to see
that this function gives rise to a one-to-one correspondence between the elements of V and the elements
of Rn . Furthermore, note that
[u + v]B = [u]B + [v]B
and
[cu]B = c[u]B ,
so that φ(u + v) = φ(u) + φ(v) and φ(cu) = cφ(u) for every u, v ∈ V and c ∈ R. Thus φ is a linear
transformation, and preserves much of the structure of V . We also say that V is isomorphic to Rn . In
practice, once we have made this identification between vectors and their coordinate matrices, then we
can basically forget about the basis B and imagine that we are working in Rn with the standard basis.
Clearly, if we change from one basis B = {u1 , . . . , un } to another basis C = {v1 , . . . , vn } of V , then we
also need to find a way of calculating [u]C in terms of [u]B for every vector u ∈ V . To do this, note that
each of the vectors v1 , . . . , vn can be written uniquely as a linear combination of the vectors u1 , . . . , un .
Suppose that for i = 1, . . . , n, we have
vi = a1i u1 + . . . + ani un ,
Chapter 8 : Linear Transformations
where a1i , . . . , ani ∈ R,
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so that


a1i
.
[vi ]B =  ..  .
ani
For every u ∈ V , we can write
u = β1 u1 + . . . + βn un = γ1 v1 + . . . + γn vn ,
where β1 , . . . , βn , γ1 , . . . , γn ∈ R,
so that



β1
 
[u]B =  ... 
and
βn

γ1
 
[u]C =  ...  .
γn
Clearly
u = γ1 v1 + . . . + γn vn
= γ1 (a11 u1 + . . . + an1 un ) + . . . + γn (a1n u1 + . . . + ann un )
= (γ1 a11 + . . . + γn a1n )u1 + . . . + (γ1 an1 + . . . + γn ann )un
= β1 u1 + . . . + βn un .
Hence
β1 = γ1 a11 + . . . + γn a1n ,
..
.
βn = γ1 an1 + . . . + γn ann .
Written in matrix notation, we have

 
β1
a11
 ..   ..
 . =
.
βn
an1
...
...
 
γ1
a1n
..   .. 
 . .
.
ann
γn
We have proved the following result.
PROPOSITION 8J. Suppose that B = {u1 , . . . , un } and C = {v1 , . . . , vn } are two bases of a real
vector space V . Then for every u ∈ V , we have
[u]B = P [u]C ,
where the columns of the matrix
P = ( [v1 ]B
...
[vn ]B )
are precisely the coordinate matrices of the elements of C relative to the basis B.
Remark. Strictly speaking, Proposition 8J gives [u]B in terms of [u]C . However, note that the matrix
P is invertible (why?), so that [u]C = P −1 [u]B .
Definition. The matrix P in Proposition 8J is sometimes called the transition matrix from the basis C
to the basis B.
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Example 8.5.2. We know that with
u1 = (1, 2, 1, 0),
u2 = (3, 3, 3, 0),
u3 = (2, −10, 0, 0),
u4 = (−2, 1, −6, 2),
and with
v1 = (1, 2, 1, 0),
v2 = (1, −1, 1, 0),
v3 = (1, 0, −1, 0),
v4 = (0, 0, 0, 2),
both B = {u1 , u2 , u3 , u4 } and C = {v1 , v2 , v3 , v4 } are bases of R4 . It is easy to check that
v1 = u1 ,
v2 = −2u1 + u2 ,
v3 = 11u1 − 4u2 + u3 ,
v4 = −27u1 + 11u2 − 2u3 + u4 ,
so that

P = ( [v1 ]B
[v2 ]B
[v3 ]B
1
0
[v4 ]B ) = 
0
0
−2
1
0
0
11
−4
1
0

−27
11 
.
−2
1
−3
4
1
0

−1
−3 
.
2
1
Hence [u]B = P [u]C for every u ∈ R4 . It is also easy to check that
u1 = v1 ,
u2 = 2v1 + v2 ,
u3 = −3v1 + 4v2 + v3 ,
u4 = −v1 − 3v2 + 2v3 + v4 ,
so that

Q = ( [u1 ]C
[u2 ]C
[u3 ]C
1
0
[u4 ]C ) = 
0
0
2
1
0
0
Hence [u]C = Q[u]B for every u ∈ R4 . Note that P Q = I. Now let u = (6, −1, 2, 2). We can check that
u = v1 + 3v2 + 2v3 + v4 , so that
 
1
3
[u]C =   .
2
1
Then

1
0
[u]B = 
0
0
−2
1
0
0
11
−4
1
0
  

−27
1
−10
11   3   6 
.
  = 
2
0
−2
1
1
1
Check that u = −10u1 + 6u2 + u4 .
Chapter 8 : Linear Transformations
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Example 8.5.3. Consider the vector space P2 . It is not too difficult to check that
u2 = 1 + x2 ,
u1 = 1 + x,
u3 = x + x2
form a basis of P2 . Let u = 1 + 4x − x2 . Then u = β1 u1 + β2 u2 + β3 u3 , where
1 + 4x − x2 = β1 (1 + x) + β2 (1 + x2 ) + β3 (x + x2 ) = (β1 + β2 ) + (β1 + β3 )x + (β2 + β3 )x2 ,
so that β1 + β2 = 1, β1 + β3 = 4 and β2 + β3 = −1. Hence (β1 , β2 , β3 ) = (3, −2, 1). If we write
B = {u1 , u2 , u3 }, then


3
[u]B =  −2  .
1
On the other hand, it is also not too difficult to check that
v1 = 1,
v2 = 1 + x,
v3 = 1 + x + x2
form a basis of P2 . Also u = γ1 v1 + γ2 v2 + γ3 v3 , where
1 + 4x − x2 = γ1 + γ2 (1 + x) + γ3 (1 + x + x2 ) = (γ1 + γ2 + γ3 ) + (γ2 + γ3 )x + γ3 x2 ,
so that γ1 + γ2 + γ3 = 1, γ2 + γ3 = 4 and γ3 = −1. Hence (γ1 , γ2 , γ3 ) = (−3, 5, −1). If we write
C = {v1 , v2 , v3 }, then


−3
[u]C =  5  .
−1
Next, note that
v1 = 21 u1 + 12 u2 − 12 u3 ,
v2 = u1 ,
v3 = 12 u1 + 12 u2 + 12 u3 .
Hence

P = ( [v1 ]B
[v2 ]B
1/2
[v3 ]B ) =  1/2
−1/2
To verify that [u]B = P [u]C , note that

 
3
1/2
 −2  =  1/2
1
−1/2

1 1/2
0 1/2  .
0 1/2


1 1/2
−3
0 1/2   5  .
0 1/2
−1
8.6. Kernel and Range
Consider first of all a euclidean linear transformation T : Rn → Rm . Suppose that A is the standard
matrix for T . Then the range of the transformation T is given by
R(T ) = {T (x) : x ∈ Rn } = {Ax : x ∈ Rn }.
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It follows that R(T ) is the set of all linear combinations of the columns of the matrix A, and is therefore
the column space of A. On the other hand, the set
{x ∈ Rn : Ax = 0}
is the nullspace of A.
Recall that the sum of the dimension of the nullspace of A and dimension of the column space of A is
equal to the number of columns of A. This is known as the Rank-nullity theorem. The purpose of this
section is to extend this result to the setting of linear transformations. To do this, we need the following
generalization of the idea of the nullspace and the column space.
Definition. Suppose that T : V → W is a linear transformation from a real vector space V into a real
vector space W . Then the set
ker(T ) = {u ∈ V : T (u) = 0}
is called the kernel of T , and the set
R(T ) = {T (u) : u ∈ V }
is called the range of T .
Example 8.6.1. For a euclidean linear transformation T with standard matrix A, we have shown that
ker(T ) is the nullspace of A, while R(T ) is the column space of A.
Example 8.6.2. Suppose that T : V → W is the zero transformation. Clearly we have ker(T ) = V and
R(T ) = {0}.
Example 8.6.3. Suppose that T : V → V is the identity operator on V . Clearly we have ker(T ) = {0}
and R(T ) = V .
Example 8.6.4. Suppose that T : R2 → R2 is orthogonal projection onto the x1 -axis. Then ker(T ) is
the x2 -axis, while R(T ) is the x1 -axis.
Example 8.6.5. Suppose that T : Rn → Rn is one-to-one. Then ker(T ) = {0} and R(T ) = Rn , in view
of Proposition 8E.
Example 8.6.6. Consider the linear transformation T : V → W , where V denotes the vector space of
all real valued functions differentiable everywhere in R, where W denotes the space of all real valued
functions defined in R, and where T (f ) = f 0 for every f ∈ V . Then ker(T ) is the set of all differentiable
functions with derivative 0, and so is the set of all constant functions in R.
Example 8.6.7. Consider the linear transformation T : V → R, where V denotes the vector space of
all real valued functions Riemann integrable over the interval [0, 1], and where
Z
T (f ) =
0
1
f (x) dx
for every f ∈ V . Then ker(T ) is the set of all Riemann integrable functions in [0, 1] with zero mean,
while R(T ) = R.
PROPOSITION 8K. Suppose that T : V → W is a linear transformation from a real vector space V
into a real vector space W . Then ker(T ) is a subspace of V , while R(T ) is a subspace of W .
Chapter 8 : Linear Transformations
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Proof. Since T (0) = 0, it follows that 0 ∈ ker(T ) ⊆ V and 0 ∈ R(T ) ⊆ W . For any u, v ∈ ker(T ), we
have
T (u + v) = T (u) + T (v) = 0 + 0 = 0,
so that u + v ∈ ker(T ). Suppose further that c ∈ R. Then
T (cu) = cT (u) = c0 = 0,
so that cu ∈ ker(T ). Hence ker(T ) is a subspace of V . Suppose next that w, z ∈ R(T ). Then there exist
u, v ∈ V such that T (u) = w and T (v) = z. Hence
T (u + v) = T (u) + T (v) = w + z,
so that w + z ∈ R(T ). Suppose further that c ∈ R. Then
T (cu) = cT (u) = cw,
so that cw ∈ R(T ). Hence R(T ) is a subspace of W . To complete this section, we prove the following generalization of the Rank-nullity theorem.
PROPOSITION 8L. Suppose that T : V → W is a linear transformation from an n-dimensional real
vector space V into a real vector space W . Then
dim ker(T ) + dim R(T ) = n.
Proof. Suppose first of all that dim ker(T ) = n. Then ker(T ) = V , and so R(T ) = {0}, and the result
follows immediately. Suppose next that dim ker(T ) = 0, so that ker(T ) = {0}. If {v1 , . . . , vn } is a
basis of V , then it follows that T (v1 ), . . . , T (vn ) are linearly independent in W , for otherwise there exist
c1 , . . . , cn ∈ R, not all zero, such that
c1 T (v1 ) + . . . + cn T (vn ) = 0,
so that T (c1 v1 + . . . + cn vn ) = 0, a contradiction since c1 v1 + . . . + cn vn 6= 0. On the other hand,
elements of R(T ) are linear combinations of T (v1 ), . . . , T (vn ). Hence dim R(T ) = n, and the result again
follows immediately. We may therefore assume that dim ker(T ) = r, where 1 ≤ r < n. Let {v1 , . . . , vr }
be a basis of ker(T ). This basis can be extended to a basis {v1 , . . . , vr , vr+1 , . . . , vn } of V . It suffices to
show that
{T (vr+1 ), . . . , T (vn )}
(4)
is a basis of R(T ). Suppose that u ∈ V . Then there exist β1 , . . . , βn ∈ R such that
u = β1 v1 + . . . + βr vr + βr+1 vr+1 + . . . + βn vn ,
so that
T (u) = β1 T (v1 ) + . . . + βr T (vr ) + βr+1 T (vr+1 ) + . . . + βn T (vn )
= βr+1 T (vr+1 ) + . . . + βn T (vn ).
It follows that (4) spans R(T ). It remains to prove that its elements are linearly independent. Suppose
that cr+1 , . . . , cn ∈ R and
cr+1 T (vr+1 ) + . . . + cn T (vn ) = 0.
Chapter 8 : Linear Transformations
(5)
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We need to show that
cr+1 = . . . = cn = 0.
(6)
By linearity, it follows from (5) that T (cr+1 vr+1 + . . . + cn vn ) = 0, so that
cr+1 vr+1 + . . . + cn vn ∈ ker(T ).
Hence there exist c1 , . . . , cr ∈ R such that
cr+1 vr+1 + . . . + cn vn = c1 v1 + . . . + cr vr ,
so that
c1 v1 + . . . + cr vr − cr+1 vr+1 − . . . − cn vn = 0.
Since {v1 , . . . , vn } is a basis of V , it follows that c1 = . . . = cr = cr+1 = . . . = cn = 0, so that (6) holds.
This completes the proof. Remark. We sometimes say that dim R(T ) and dim ker(T ) are respectively the rank and the nullity of
the linear transformation T .
8.7. Inverse Linear Transformations
In this section, we generalize some of the ideas first discussed in Section 8.3.
Definition. A linear transformation T : V → W from a real vector space V into a real vector space W
is said to be one-to-one if for every u0 , u00 ∈ V , we have u0 = u00 whenever T (u0 ) = T (u00 ).
The result below follows immediately from our definition.
PROPOSITION 8M. Suppose that T : V → W is a linear transformation from a real vector space V
into a real vector space W . Then T is one-to-one if and only if ker(T ) = {0}.
Proof. (⇒) Clearly 0 ∈ ker(T ). Suppose that ker(T ) 6= {0}. Then there exists a non-zero v ∈ ker(T ).
It follows that T (v) = T (0), and so T is not one-to-one.
(⇐) Suppose that ker(T ) = {0}. Given any u0 , u00 ∈ V , we have
T (u0 ) − T (u00 ) = T (u0 − u00 ) = 0
if and only if u0 − u00 = 0; in other words, if and only if u0 = u00 . We have the following generalization of Proposition 8E.
PROPOSITION 8N. Suppose that T : V → V is a linear operator on a finite-dimensional real vector
space V . Then the following statements are equivalent:
(a) The linear operator T is one-to-one.
(b) We have ker(T ) = {0}.
(c) The range of T is V ; in other words, R(T ) = V .
Proof. The equivalence of (a) and (b) is established by Proposition 8M. The equivalence of (b) and (c)
follows from Proposition 8L. Chapter 8 : Linear Transformations
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Suppose that T : V → W is a one-to-one linear transformation from a real vector space V into a real
vector space W . Then for every w ∈ R(T ), there exists exactly one u ∈ V such that T (u) = w. We can
therefore define a transformation T −1 : R(T ) → V by writing T −1 (w) = u, where u ∈ V is the unique
vector satisfying T (u) = w.
PROPOSITION 8P. Suppose that T : V → W is a one-to-one linear transformation from a real vector
space V into a real vector space W . Then T −1 : R(T ) → V is a linear transformation.
Proof. Suppose that w, z ∈ R(T ). Then there exist u, v ∈ V such that T −1 (w) = u and T −1 (z) = v.
It follows that T (u) = w and T (v) = z, so that T (u + v) = T (u) + T (v) = w + z, whence
T −1 (w + z) = u + v = T −1 (w) + T −1 (z).
Suppose further that c ∈ R. Then T (cu) = cw, so that
T −1 (cw) = cu = cT −1 (w).
This completes the proof. We also have the following result concerning compositions of linear transformations and which requires
no further proof, in view of our knowledge concerning inverse functions.
PROPOSITION 8Q. Suppose that V, W, U are real vector spaces. Suppose further that T1 : V → W
and T2 : W → U are one-to-one linear transformations. Then
(a) the linear transformation T2 ◦ T1 : V → U is one-to-one; and
(b) (T2 ◦ T1 )−1 = T1−1 ◦ T2−1 .
8.8. Matrices of General Linear Transformations
Suppose that T : V → W is a linear transformation from a real vector space V to a real vector space W .
Suppose further that the vector spaces V and W are finite dimensional, with dim V = n and dim W = m.
We shall show that if we make use of a basis B of V and a basis C of W , then it is possible to describe
T indirectly in terms of some matrix A. The main idea is to make use of coordinate matrices relative to
the bases B and C.
Let us recall some discussion in Section 8.5. Suppose that B = {v1 , . . . , vn } is a basis of V . Then
every vector v ∈ V can be written uniquely as a linear combination
v = β1 v 1 + . . . + βn v n ,
The matrix
where β1 , . . . , βn ∈ R.

β1
 . 
[v]B =  .. 
(7)

(8)
βn
is the coordinate matrix of v relative to the basis B.
Consider now a transformation φ : V → Rn , where φ(v) = [v]B for every v ∈ V . The proof of the
following result is straightforward.
PROPOSITION 8R. Suppose that the real vector space V has basis B = {v1 , . . . , vn }. Then the
transformation φ : V → Rn , where φ(v) = [v]B satisfies (7) and (8) for every v ∈ V , is a oneto-one linear transformation, with range R(φ) = Rn . Furthermore, the inverse linear transformation
φ−1 : Rn → V is also one-to-one, with range R(φ−1 ) = V .
Chapter 8 : Linear Transformations
page 20 of 35
c W W L Chen, 1997, 2006
!
!
W W L Chen, 1997, 2006
cc W
2008
!
W L Chen, 1997, 2006
Linear Algebra
Linear Algebra
Linear Algebra
Suppose
next
that
=
{w
w
} is
basis
of
W
Then
we
can
define
linear
transformation
Suppose next
next
that CC
C=
= {w
{w11,,, ... ... ... ,,, w
wm
is aaa basis
basis of
of W
W... Then
Then we
we can
can define
define aaa linear
linear transformation
transformation
m } is
m that
Suppose
1 [w]C for
m } every
ψ
:
W
→
R
w
∈
W
,
in
a
similar
way.
We
now
have
the following
m , where ψ(w) =
ψ
:
W
→
R
m , where ψ(w) = [w]C for every w ∈ W , in a similar way. We now have the following
ψ
: W →ofRlinear
, where
ψ(w) = [w]C for every w ∈ W , in a similar way. We now have the following
diagram
transformations.
diagram of
of linear
linear transformations.
transformations.
diagram
T
T
T
V
VV
φ−1
φ−1
φ−1
ψ −1
ψ −1
ψ −1
φ
φ
φ
n
R
Rnn
R
Clearly
Clearly the
the composition
composition
Clearly
Clearly the
the composition
composition
W
W
W
ψ
ψ
ψ
m
R
m
Rm
R
−1
n
m
−1
n
m
S=
=
ψ
T
φ
R
→
R
−1 :: R
m
S
ψ
T
φ
→
R
=ψ
ψ ◦◦◦◦ T
T ◦◦◦◦ φ
φ−1
Rnn →
→R
Rm
SS =
:: R
is a euclidean linear transformation, and can therefore be described in terms of a standard matrix A.
is aa euclidean
euclidean linear
linear transformation,
transformation, and
and can therefore
therefore be
be described
described in terms
terms of
of aa standard
standard matrix
matrix A.
A.
is
Our
task is to determine
this matrix A in can
terms of T and
the bases Binand
C.
Our
task
is
to
determine
this
matrix
A
in
terms
of
T
and
the
bases
B
and
C.
Our task is to determine this matrix A in terms of T and the bases B and C.
We know from Proposition 8A that
We know
know from
from Proposition
Proposition 8A
8A that
that
We
A = ( S(e1 ) . . . S(en ) ) ,
A=
= (( S(e
S(e1)) .. .. .. S(e
S(en)) )) ,,
A
1
n
where {e1 , . . . , en } is the standard basis for Rnn. For every j = 1, . . . , n, we have
where {e
{e1,, .. .. .. ,, een}} is
is the
the standard
standard basis
basis for
for R
Rn .. For
For every
every jj =
= 1,
1, .. .. .. ,, n,
n, we
we have
have
where
1
n
−1
−1
S(ej ) = (ψ ◦ T ◦ φ −1)(ej ) = ψ(T (φ −1(ej ))) = ψ(T (vj )) = [T (vj )]C .
S(ej)) =
= (ψ
(ψ ◦◦ TT ◦◦ φφ−1 )(e
)(ej)) =
= ψ(T
ψ(T(φ
(φ−1 (e
(ej)))
))) =
= ψ(T
ψ(T(v
(vj))
)) =
= [T
[T(v
(vj)]
)]C .
S(e
j
j
j
j
j C.
It follows that
It follows
follows that
that
It
A = ( [T (v1 )]C . . . [T (vn )]C ) .
(9)
(9)
)]C .. .. .. [T
[T(v
(vn)]
)]C ) .
(9)
A=
= (( [T
[T(v
(v1)]
(9)
A
1 C
n C ).
Definition.
The
matrix
A
given
by
(9)
is
called
the
matrix
for
the
linear
transformation
T
with
respect
Definition. The
The matrix
matrix A
A given
given by
by (9)
(9) is
is called
called the
the matrix
matrix for
for the
the linear
linear transformation
transformation T
T with
with respect
respect
Definition.
to
the
bases
B
and
C.
to
the
bases
B
and
C.
to the bases B and C.
We
now
have
the
following
diagram
of
linear
transformations.
We now
now have
have the
the following
following diagram
diagram of
of linear
linear transformations.
transformations.
We
T
T
T
V
VV
φ−1
φ−1
φ−1
W
W
W
ψ −1
ψ −1
ψ −1
φ
φ
φ
ψ
ψ
ψ
S
n
m
S
R
R
m
S
Rnn
Rm
R
R
Hence we
we can
can write
write T
T as
as the
the composition
composition
Hence
Hence
we can
write T
as the
composition
−1
T=
=ψ
ψ−1
◦ S ◦ φ : V → W.
T
−1 ◦ S ◦ φ : V → W.
T
=ψ
◦ S ◦ φ : V → W.
For every v ∈ V , we have the following:
For every
every v
v∈
∈V
V ,, we
we have
have the
the following:
following:
For
v
vv
φ
φ
φ
Chapter 8 : Linear Transformations
Chapter 8 : Linear Transformations
Chapter 8 : Linear Transformations
[v]
[v]B
B
[v]
B
S
S
S
A[v]
A[v]B
B
A[v]
B
ψ −1
ψ −1
ψ −1
−1
ψ
−1 (A[v]B )
ψ−1
(A[v]B))
ψ
(A[v]
B
page 21 of 35
page 21 of 35
page 21 of 35
c
Linear Algebra
W W L Chen, 1997, 2008
More precisely, if v = β1 v1 + . . . + βn vn , then


β1
 . 
[v]B =  .. 

and
βn
 
β1
 .  
A[v]B = A  ..  = 
βn

γ1
..  ,
. 
γm
say, and so T (v) = ψ −1 (A[v]B ) = γ1 w1 + . . . + γm wm . We have proved the following result.
PROPOSITION 8S. Suppose that T : V → W is a linear transformation from a real vector space V
into a real vector space W . Suppose further that V and W are finite dimensional, with bases B and C
respectively, and that A is the matrix for the linear transformation T with respect to the bases B and C.
Then for every v ∈ V , we have T (v) = w, where w ∈ W is the unique vector satisfying [w]C = A[v]B .
Remark. In the special case when V = W , the linear transformation T : V → W is a linear operator
on T . Of course, we may choose a basis B for the domain V of T and a basis C for the codomain V
of T . In the case when T is the identity linear operator, we often choose B 6= C since this represents a
change of basis. In the case when T is not the identity operator, we often choose B = C for the sake of
convenience; we then say that A is the matrix for the linear operator T with respect to the basis B.
Example 8.8.1. Consider an operator T : P3 → P3 on the real vector space P3 of all polynomials with
real coefficients and degree at most 3, where for every polynomial p(x) in P3 , we have T (p(x)) = xp0 (x),
the product of x with the formal derivative p0 (x) of p(x). The reader is invited to check that T is a
linear operator. Now consider the basis B = {1, x, x2 , x3 } of P3 . The matrix for T with respect to B is
given by

A = ( [T (1)]B
[T (x)]B
[T (x2 )]B
[T (x3 )]B ) = ( [0]B
[x]B
[2x2 ]B
0
0
3
[3x ]B ) = 
0
0
0
1
0
0
0
0
2
0

0
0
.
0
3
Suppose that p(x) = 1 + 2x + 4x2 + 3x3 . Then
 
1
2
[p(x)]B =  
4
3

and
0
0
A[p(x)]B = 
0
0
0
1
0
0
0
0
2
0
   
0
1
0
02 2
  =  ,
0
4
8
3
3
9
so that T (p(x)) = 2x + 8x2 + 9x3 . This can be easily verified by noting that
T (p(x)) = xp0 (x) = x(2 + 8x + 9x2 ) = 2x + 8x2 + 9x3 .
In general, if p(x) = p0 + p1 x + p2 x2 + p3 x3 , then


p0
p 
[p(x)]B =  1 
p2
p3

and
0
0
A[p(x)]B = 
0
0
0
1
0
0
0
0
2
0
  

0
p0
0
0   p1   p1 
  = 
,
0
p2
2p2
3
p3
3p3
so that T (p(x)) = p1 x + 2p2 x2 + 3p3 x3 . Observe that
T (p(x)) = xp0 (x) = x(p1 + 2p2 x + 3p3 x2 ) = p1 x + 2p2 x2 + 3p3 x3 ,
verifying our result.
Chapter 8 : Linear Transformations
page 22 of 35
c W
!
WW
WL
L Chen,
Chen, 1997,
1997, 2006
2008
Linear
Linear Algebra
Algebra
Example 8.8.2. Consider the linear operator T : R22 → R22, given by T (x11, x22) = (2x11 + x22, x11 + 3x22)
for every (x11, x22) ∈ R22. Consider also the basis B = {(1, 0), (1, 1)} of R22. Then the matrix for T with
respect to B is given by
!
A = ( [T (1, 0)]BB [T (1, 1)]BB ) = ( [(2, 1)]BB [(3, 4)]BB ) =
"
1 −1
.
1 4
Suppose that (x11, x22) = (3, 2). Then
[(3, 2)]BB =
!
"
1
2
!
and
A[(3, 2)]BB =
1 −1
1 4
"
"
!
"
!
1
−1
=
,
2
9
so that T (3, 2) = −(1, 0) + 9(1, 1) = (8, 9). This can be easily verified directly. In general, we have
!
[(x11, x22)]BB =
x11 − x22
x22
"
!
and
A[(x11, x22)]BB =
1 −1
1 4
"
!
x11 − x22
x22
"
!
=
x11 − 2x22
x11 + 3x22
"
,
so that T (x11, x22) = (x11 − 2x22)(1, 0) + (x11 + 3x22)(1, 1) = (2x11 + x22, x11 + 3x22).
m
Example 8.8.3. Suppose that T : Rnn → Rm
is a linear transformation. Suppose further that B and C
n
m
n
m
are the standard bases for R and R respectively. Then the matrix for T with respect to B and C is
given by
A = ( [T (e11)]CC . . . [T (enn)]CC ) = ( T (e11) . . . T (enn) ) ,
so it follows from Proposition 8A that A is simply the standard matrix for T .
Suppose now that T11 : V → W and T22 : W → U are linear transformations, where the real vector
spaces V, W, U are finite dimensional, with respective bases B = {v11, . . . , vnn}, C = {w11, . . . , wm
m} and
D = {u11, . . . , ukk}. We then have the following diagram of linear transformations.
T1
V
φ−1
ψ −1
φ
Rn
S1
T2
W
η −1
ψ
Rm
U
S2
η
Rk
Here η : U → Rkk, where η(u) = [u]D
D for every u ∈ U , is a linear transformation, and
−1 : Rn
n → Rm
m
S11 = ψ ◦ T11 ◦ φ−1
and
−1 : Rm
m → Rkk
S22 = η ◦ T22 ◦ ψ −1
are euclidean linear transformations. Suppose that A11 and A22 are respectively the standard matrices
for S11 and S22, so that they are respectively the matrix for T11 with respect to B and C and the matrix
for T22 with respect to C and D. Clearly
−1 : Rn
n → Rkk.
S22 ◦ S11 = η ◦ T22 ◦ T11 ◦ φ−1
It follows that A22A11 is the standard matrix for S22 ◦ S11, and so is the matrix for T22 ◦ T11 with respect to
the bases B and D. To summarize, we have the following result.
Chapter
Chapter 88 :: Linear
Linear Transformations
Transformations
page
page 23
23 of
of 35
35
c
Linear Algebra
W W L Chen, 1997, 2008
PROPOSITION 8T. Suppose that T1 : V → W and T2 : W → U are linear transformations, where
the real vector spaces V, W, U are finite dimensional, with bases B, C, D respectively. Suppose further that
A1 is the matrix for the linear transformation T1 with respect to the bases B and C, and that A2 is the
matrix for the linear transformation T2 with respect to the bases C and D. Then A2 A1 is the matrix for
the linear transformation T2 ◦ T1 with respect to the bases B and D.
Example 8.8.4. Consider the linear operator T1 : P3 → P3 , where for every polynomial p(x) in P3 ,
we have T1 (p(x)) = xp0 (x). We have already shown that the matrix for T1 with respect to the basis
B = {1, x, x2 , x3 } of P3 is given by

0
0
A1 = 
0
0
0
1
0
0
0
0
2
0

0
0
.
0
3
Consider next the linear operator T2 : P3 → P3 , where for every polynomial q(x) = q0 + q1 x + q2 x2 + q3 x3
in P3 , we have
T2 (q(x)) = q(1 + x) = q0 + q1 (1 + x) + q2 (1 + x)2 + q3 (1 + x)3 .
We have T2 (1) = 1, T2 (x) = 1 + x, T2 (x2 ) = 1 + 2x + x2 and T2 (x3 ) = 1 + 3x + 3x2 + x3 , so that the
matrix for T2 with respect to B is given by

A2 = ( [T2 (1)]B
[T2 (x)]B
1
0
3
[T2 (x )]B ) = 
0
0
[T2 (x2 )]B
1
1
0
0
1
2
1
0

1
3
.
3
1
Consider now the composition T = T2 ◦ T1 : P3 → P3 . Let A denote the matrix for T with respect to B.
By Proposition 8T, we have

1
0
A = A2 A1 = 
0
0
1
1
0
0
1
2
1
0

1
0
30

3
0
1
0
0
1
0
0
 
0
0
0 0
=
0
0
3
0
0
0
2
0
1
1
0
0
2
4
2
0

3
9
.
9
3
Suppose that p(x) = p0 + p1 x + p2 x2 + p3 x3 . Then


p0
p 
[p(x)]B =  1 
p2
p3

and
0
0
A[p(x)]B = 
0
0
1
1
0
0
2
4
2
0
  

3
p0
p1 + 2p2 + 3p3
9   p1   p1 + 4p2 + 9p3 
  = 
,
9
p2
2p2 + 9p3
3
p3
3p3
so that T (p(x)) = (p1 + 2p2 + 3p3 ) + (p1 + 4p2 + 9p3 )x + (2p2 + 9p3 )x2 + 3p3 x3 . We can check this directly
by noting that
T (p(x)) = T2 (T1 (p(x))) = T2 (p1 x + 2p2 x2 + 3p3 x3 ) = p1 (1 + x) + 2p2 (1 + x)2 + 3p3 (1 + x)3
= (p1 + 2p2 + 3p3 ) + (p1 + 4p2 + 9p3 )x + (2p2 + 9p3 )x2 + 3p3 x3 .
Example 8.8.5. Consider the linear operator T : R2 → R2 , given by T (x1 , x2 ) = (2x1 + x2 , x1 + 3x2 )
for every (x1 , x2 ) ∈ R2 . We have already shown that the matrix for T with respect to the basis
B = {(1, 0), (1, 1)} of R2 is given by
A=
Chapter 8 : Linear Transformations
1
1
−1
4
.
page 24 of 35
c
Linear Algebra
W W L Chen, 1997, 2008
Consider the linear operator T 2 : R2 → R2 . By Proposition 8T, the matrix for T 2 with respect to B is
given by
1 −1
1 −1
0 −5
2
A =
=
.
1 4
1 4
5 15
Suppose that (x1 , x2 ) ∈ R2 . Then
x1 − x2
[(x1 , x2 )]B =
and
x2
2
A [(x1 , x2 )]B =
0
5
−5
15
x1 − x2
x2
=
−5x2
5x1 + 10x2
,
so that T (x1 , x2 ) = −5x2 (1, 0) + (5x1 + 10x2 )(1, 1) = (5x1 + 5x2 , 5x1 + 10x2 ). The reader is invited to
check this directly.
A simple consequence of Propositions 8N and 8T is the following result concerning inverse linear
transformations.
PROPOSITION 8U. Suppose that T : V → V is a linear operator on a finite dimensional real vector
space V with basis B. Suppose further that A is the matrix for the linear operator T with respect to the
basis B. Then T is one-to-one if and only if A is invertible. Furthermore, if T is one-to-one, then A−1
is the matrix for the inverse linear operator T −1 : V → V with respect to the basis B.
Proof. Simply note that T is one-to-one if and only if the system Ax = 0 has only the trivial solution
x = 0. The last assertion follows easily from Proposition 8T, since if A0 denotes the matrix for the
inverse linear operator T −1 with respect to B, then we must have A0 A = I, the matrix for the identity
operator T −1 ◦ T with respect to B. Example 8.8.6. Consider the linear operator T : P3 → P3 , where for every q(x) = q0 + q1 x + q2 x2 + q3 x3
in P3 , we have
T (q(x)) = q(1 + x) = q0 + q1 (1 + x) + q2 (1 + x)2 + q3 (1 + x)3 .
We have already shown that the matrix for T with respect to the basis B = {1, x, x2 , x3 } is given by

1
0
A=
0
0
1
1
0
0
1
2
1
0

1
3
.
3
1
This matrix is invertible, so it follows that T is one-to-one. Furthermore, it can be checked that

A−1
1
0
=
0
0
−1
1
0
0

1 −1
−2 3 
.
1 −3
0
1
Suppose that p(x) = p0 + p1 x + p2 x2 + p3 x3 . Then


p0
p 
[p(x)]B =  1 
p2
p3

and
1
0

A−1 [p(x)]B = 
0
0
−1
1
0
0
1
−2
1
0
  

−1
p0
p0 − p1 + p2 − p3
3   p1   p1 − 2p2 + 3p3 
  = 
,
−3
p2
p2 − 3p3
1
p3
p3
so that
T −1 (p(x)) = (p0 − p1 + p2 − p3 ) + (p1 − 2p2 + 3p3 )x + (p2 − 3p3 )x2 + p3 x3
= p0 + p1 (x − 1) + p2 (x2 − 2x + 1) + p3 (x3 − 3x2 + 3x − 1)
= p0 + p1 (x − 1) + p2 (x − 1)2 + p3 (x − 1)3 = p(x − 1).
Chapter 8 : Linear Transformations
page 25 of 35
c
Linear Algebra
W W L Chen, 1997, 2008
8.9. Change of Basis
Suppose that V is a finite dimensional real vector space, with one basis B = {v1 , . . . , vn } and another
basis B 0 = {u1 , . . . , un }. Suppose that T : V → V is a linear operator on V . Let A denote the matrix
for T with respect to the basis B, and let A0 denote the matrix for T with respect to the basis B 0 . If
v ∈ V and T (v) = w, then
[w]B = A[v]B
(10)
[w]B0 = A0 [v]B0 .
(11)
and
We wish to find the relationship between A0 and A.
Recall Proposition 8J, that if
P = ( [u1 ]B
...
[un ]B )
denotes the transition matrix from the basis B 0 to the basis B, then
[v]B = P [v]B0
and
[w]B = P [w]B0 .
(12)
Note that the matrix P can also be interpreted as the matrix for the identity operator I : V → V with
respect to the bases B 0 and B. It is easy to see that the matrix P is invertible, and
P −1 = ( [v1 ]B0
...
[vn ]B0 )
denotes the transition matrix from the basis B to the basis B 0 , and can also be interpreted as the matrix
for the identity operator I : V → V with respect to the bases B and B 0 .
Combining (10) and (12), we conclude that
[w]B0 = P −1 [w]B = P −1 A[v]B = P −1 AP [v]B0 .
Comparing this with (11), we conclude that
P −1 AP = A0 .
(13)
A = P A0 P −1 .
(14)
This implies that
Remark. We can use the notation
A = [T ]B
and
A0 = [T ]B0
to denote that A and A0 are the matrices for T with respect to the basis B and with respect to the basis
B 0 respectively. We can also write
P = [I]B,B0
to denote that P is the transition matrix from the basis B 0 to the basis B, so that
P −1 = [I]B0 ,B .
Chapter 8 : Linear Transformations
page 26 of 35
cc
!
Linear
Linear Algebra
Algebra
W
WW
WL
L Chen,
Chen, 1997,
1997, 2006
2008
Then
Then (13)
(13) and
and (14)
(14) become
become respectively
respectively
!
! = [T ] !0
[I]
[T ]B [I]
[I]B
[I]B,B
B0 ,B
,B [T ]B
B,B0 = [T ]B
B
and
and
! [T ] !0 [I]B !0 ,B = [T ]B .
[I]
[I]B,B
B,B0 [T ]B
B [I]B ,B = [T ]B .
We have proved the following result.
PROPOSITION 8V. Suppose that T : V → V is a linear operator on a finite dimensional
space V , with bases B = {v11 , . . . , vnn } and B !0 = {u11 , . . . , unn }. Suppose further that A and
matrices for T with respect to the basis B and with respect to the basis B !0 respectively. Then
P −1 AP = A!0
real vector
A!0 are the
A!0 = P AP −1 ,
and
where
P = ( [u1 ]B
...
[un ]B )
denotes the transition matrix from the basis B !0 to the basis B.
Remarks. (1) We have the following picture.
T
v
w
I
I
v
w
T
A!
[v]B!
[w]B!
P
P −1
[v]B
A
[w]B
(2)
(2) The
The idea
idea can
can be
be extended
extended to
to the
the case
case of
of linear
linear transformations
transformations T
T :: V
V →
→W
W from
from aa finite
finite dimensional
dimensional
real
vector
space
into
another,
with
a
change
of
basis
in
V
and
a
change
of
basis
in
W
real vector space into another, with a change of basis in V and a change of basis in W ..
Example
Example 8.9.1.
8.9.1. Consider
Consider the
the vector
vector space
space P
P33 of
of all
all polynomials
polynomials with
with real
real coefficients
coefficients and
and degree
degree at
at
2
2 , x3
3 } and B !0 = {1, 1 + x, 1 + x + x2
2 , 1 + x + x2
2 + x3
3 }. Consider also
most
3,
with
bases
B
=
{1,
x,
x
most 3, with bases B = {1, x, x , x } and B = {1, 1 + x, 1 + x + x , 1 + x + x + x }. Consider also
2
3
the
the linear
linear operator
operator T
T :: P
P33 →
→P
P33 ,, where
where for
for every
every polynomial
polynomial p(x)
p(x) =
= pp00 +
+ pp11 x
x+
+ pp22 x
x2 +
+ pp33 x
x3 ,, we
we have
have
2
3
2 + (p0 + p3 )x3 . Let A denote the matrix for T with respect
T
(p(x))
=
(p
+
p
)
+
(p
+
p
)x
+
(p
+
p
)x
0
1
1
2
2
3
T (p(x)) = (p0 + p1 ) + (p1 + p2 )x + (p2 + p3 )x + (p0 + p3 )x . Let A denote the matrix for T with respect
3
2
2
3
2
3
to
to the
the basis
basis B.
B. Then
Then T
T (1)
(1) =
= 11 +
+x
x3 ,, T
T (x)
(x) =
= 11 +
+ x,
x, T
T (x
(x2 )) =
=x
x+
+x
x2 and
and T
T (x
(x3 )) =
=x
x2 +
+x
x3 ,, and
and so
so
A
A=
= (( [T
[T (1)]
(1)]B
B
[T
[T (x)]
(x)]B
B
2
[T
[T (x
(x2 )]
)]B
B

1
1
00

3

[T
)=
[T (x
(x3 )]
)]B
B ) = 0
0
11
11
11
00
00
00
11
11
00

00 
00 
.

11  .
11
!
Next,
Next, note
note that
that the
the transition
transition matrix
matrix from
from the
the basis
basis B
B 0 to
to the
the basis
basis B
B is
is given
given by
by

1 1
1 1
0

2
2
3
0 11
P
[1
[1
[1
)=
P =
= (( [1]
[1]B
[1 +
+ x]
x]B
[1 +
+x
x+
+x
x2 ]]B
[1 +
+x
x+
+x
x2 +
+x
x3 ]]B
0 0
B
B
B
B)=
0 0
00 00
Chapter
Chapter 8
8 :: Linear
Linear Transformations
Transformations
11
11
11
00

11 
11 
.

11  .
11
page
page 27
27 of
of 35
35
c
Linear Algebra
W W L Chen, 1997, 2008
It can be checked that

P −1
1
0
=
0
0
−1
1
0
0
0
−1
1
0

0
0 
,
−1
1
and so

1
0
0
−1
A = P AP = 
0
0
−1
1
0
0
0
−1
1
0

0
1
0 0

−1
0
1
1
1
1
0
0
0
1
1
0

0
1
00

1
0
1
0
1
1
0
0
1
1
1
0
 
1
1
1  0
=
−1
1
1
1
1
1
−1
1

0 0
1 0

0 0
1 2
is the matrix for T with respect to the basis B 0 . It follows that
T (1) = 1 − (1 + x + x2 ) + (1 + x + x2 + x3 ) = 1 + x3 ,
T (1 + x) = 1 + (1 + x) − (1 + x + x2 ) + (1 + x + x2 + x3 ) = 2 + x + x3 ,
T (1 + x + x2 ) = (1 + x) + (1 + x + x2 + x3 ) = 2 + 2x + x2 + x3 ,
T (1 + x + x2 + x3 ) = 2(1 + x + x2 + x3 ) = 2 + 2x + 2x2 + 2x3 .
These can be verified directly.
8.10. Eigenvalues and Eigenvectors
Definition. Suppose that T : V → V is a linear operator on a finite dimensional real vector space V .
Then any real number λ ∈ R is called an eigenvalue of T if there exists a non-zero vector v ∈ V such that
T (v) = λv. This non-zero vector v ∈ V is called an eigenvector of T corresponding to the eigenvalue λ.
The purpose of this section is to show that the problem of eigenvalues and eigenvectors of the linear
operator T can be reduced to the problem of eigenvalues and eigenvectors of the matrix for T with
respect to any basis B of V . The starting point of our argument is the following theorem, the proof of
which is left as an exercise.
PROPOSITION 8W. Suppose that T : V → V is a linear operator on a finite dimensional real vector
space V , with bases B and B 0 . Suppose further that A and A0 are the matrices for T with respect to the
basis B and with respect to the basis B 0 respectively. Then
(a) det A = det A0 ;
(b) A and A0 have the same rank;
(c) A and A0 have the same characteristic polynomial;
(d) A and A0 have the same eigenvalues; and
(e) the dimension of the eigenspace of A corresponding to an eigenvalue λ is equal to the dimension of
the eigenspace of A0 corresponding to λ.
We also state without proof the following result.
PROPOSITION 8X. Suppose that T : V → V is a linear operator on a finite dimensional real vector
space V . Suppose further that A is the matrix for T with respect to a basis B of V . Then
(a) the eigenvalues of T are precisely the eigenvalues of A; and
(b) a vector u ∈ V is an eigenvector of T corresponding to an eigenvalue λ if and only if the coordinate
matrix [u]B is an eigenvector of A corresponding to the eigenvalue λ.
Chapter 8 : Linear Transformations
page 28 of 35
c
Linear Algebra
W W L Chen, 1997, 2008
Suppose now that A is the matrix for a linear operator T : V → V on a finite dimensional real
vector space V with respect to a basis B = {v1 , . . . , vn }. If A can be diagonalized, then there exists an
invertible matrix P such that
P −1 AP = D
is a diagonal matrix. Furthermore, the columns of P are eigenvectors of A, and so are the coordinate
matrices of eigenvectors of T with respect to the basis B. In other words,
P = ( [u1 ]B
...
[un ]B ) ,
where B 0 = {u1 , . . . , un } is a basis of V consiting of eigenvectors of T . Furthermore, P is the transition
matrix from the basis B 0 to the basis B. It follows that the matrix for T with respect to the basis B 0 is
given by

D=

λ1
..
,
.
λn
where λ1 , . . . , λn are the eigenvalues of T .
Example 8.10.1. Consider the vector space P2 of all polynomials with real coefficients and degree at
most 2, with basis B = {1, x, x2 }. Consider also the linear operator T : P2 → P2 , where for every
polynomial p(x) = p0 + p1 x + p2 x2 , we have T (p(x)) = (5p0 − 2p1 ) + (6p1 + 2p2 − 2p0 )x + (2p1 + 7p2 )x2 .
Then T (1) = 5 − 2x, T (x) = −2 + 6x + 2x2 and T (x2 ) = 2x + 7x2 , so that the matrix for T with respect
to the basis B is given by

A = ( [T (1)]B
[T (x)]B
5
[T (x2 )]B ) =  −2
0
−2
6
2

0
2.
7
It is a simple exercise to show that the matrix A has eigenvalues 3, 6, 9, with corresponding eigenvectors


2
x1 =  2  ,
−1

−1
x3 =  2  ,
2


2
x2 =  −1  ,
2

so that writing

2
P = 2
−1

−1
2 ,
2
2
−1
2
we have

3
P −1 AP =  0
0
0
6
0

0
0.
9
Now let B 0 = {p1 (x), p2 (x), p3 (x)}, where


2
[p1 (x)]B =  2  ,
−1
Chapter 8 : Linear Transformations


2
[p2 (x)]B =  −1  ,
2


−1
[p3 (x)]B =  2  .
2
page 29 of 35
c
Linear Algebra
W W L Chen, 1997, 2008
Then P is the transition matrix from the basis B 0 to the basis B, and D is the matrix for T with respect
to the basis B 0 . Clearly p1 (x) = 2 + 2x − x2 , p2 (x) = 2 − x + 2x2 and p3 (x) = −1 + 2x + 2x2 . Note now
that
T (p1 (x)) = T (2 + 2x − x2 ) = 6 + 6x − 3x2 = 3p1 (x),
T (p2 (x)) = T (2 − x + 2x2 ) = 12 − 6x + 12x2 = 6p2 (x),
T (p3 (x)) = T (−1 + 2x + 2x2 ) = −9 + 18x + 18x2 = 9p3 (x).
Chapter 8 : Linear Transformations
page 30 of 35
c
Linear Algebra
W W L Chen, 1997, 2008
Problems for Chapter 8
1. Consider the transformation T : R3 → R4 , given by
T (x1 , x2 , x3 ) = (x1 + x2 + x3 , x2 + x3 , 3x1 + x2 , 2x2 + x3 )
for every (x1 , x2 , x3 ) ∈ R3 .
a) Find the standard matrix A for T .
b) By reducing A to row echelon form, determine the dimension of the kernel of T and the dimension
of the range of T .
2. Consider a linear operator T : R3 → R3 with standard matrix

1
A = 2
1

3
3.
2
2
1
3
Let {e1 , e2 , e3 } denote the standard basis for R3 .
a) Find T (ej ) for every j = 1, 2, 3.
b) Find T (2e1 + 5e2 + 3e3 ).
c) Is T invertible? Justify your assertion.
3. Consider the linear operator T : R2 → R2 with standard matrix
A=
1
0
1
1
.
a) Find the image under T of the line x1 + 2x2 = 3.
b) Find the image under T of the circle x21 + x22 = 1.
4. For each of the following, determine whether the given transformation is linear:
a) T : V → R, where V is a real inner product space and T (u) = kuk.
b) T : M2,2 (R) → M2,3 (R), where B ∈ M2,3 (R) is fixed and T (A) = AB.
c) T : M3,4 (R) → M4,3 (R), where T (A) = At .
d) T : P2 → P2 , where T (p0 + p1 x + p2 x2 ) = p0 + p1 (2 + x) + p2 (2 + x)2 .
e) T : P2 → P2 , where T (p0 + p1 x + p2 x2 ) = p0 + p1 x + (p2 + 1)x2 .
5. Suppose that T : R3 → R3 is a linear transformation satisfying the conditions T (1, 0, 0) = (2, 4, 1),
T (1, 1, 0) = (3, 0, 2) and T (1, 1, 1) = (1, 4, 6).
a) Evaluate T (5, 3, 2).
b) Find T (x1 , x2 , x3 ) for every (x1 , x2 , x3 ) ∈ R3 .
6. Suppose that T : R3 → R3 is orthogonal projection onto the x1 x2 -plane.
a) Find the standard matrix A for T .
b) Find A2 .
c) Show that T ◦ T = T .
7. Consider the bases B = {u1 , u2 , u3 } and C = {v1 , v2 , v3 } of R3 , where u1 = (2, 1, 1), u2 = (2, −1, 1),
u3 = (1, 2, 1), v1 = (3, 1, −5), v2 = (1, 1, −3) and v3 = (−1, 0, 2).
a) Find the transition matrix from the basis C to the basis B.
b) Find the transition matrix from the basis B to the basis C.
c) Show that the matrices in parts (a) and (b) are inverses of each other.
d) Compute the coordinate matrix [u]C , where u = (−5, 8, −5).
e) Use the transition matrix to compute the coordinate matrix [u]B .
f) Compute the coordinate matrix [u]B directly and compare it to your answer in part (e).
Chapter 8 : Linear Transformations
page 31 of 35
c
Linear Algebra
W W L Chen, 1997, 2008
8. Consider the bases B = {p1 , p2 } and C = {q1 , q2 } of P1 , where p1 = 2, p2 = 3 + 2x, q1 = 6 + 3x and
q2 = 10 + 2x.
a) Find the transition matrix from the basis C to the basis B.
b) Find the transition matrix from the basis B to the basis C.
c) Show that the matrices in parts (a) and (b) are inverses of each other.
d) Compute the coordinate matrix [p]C , where p = −4 + x.
e) Use the transition matrix to compute the coordinate matrix [p]B .
f) Compute the coordinate matrix [p]B directly and compare it to your answer in part (e).
9. Let V be the real vector space spanned by the functions f1 = sin x and f2 = cos x.
a) Show that g1 = 2 sin x + cos x and g2 = 3 cos x form a basis of V .
b) Find the transition matrix from the basis C = {g1 , g2 } to the basis B = {f1 , f2 } of V .
c) Compute the coordinate matrix [f ]C , where f = 2 sin x − 5 cos x.
d) Use the transition matrix to compute the coordinate matrix [f ]B .
e) Compute the coordinate matrix [f ]B directly and compare it to your answer in part (d).
10. Let P be the transition matrix from a basis C to another basis B of a real vector space V . Explain
why P is invertible.
11. For each of the following linear transformations T , find ker(T ) and R(T ), and verify the Rank-nullity
theorem:


1 −1 3
a) T : R3 → R3 , with standard matrix A =  5 6 −4 .
7 4
2
b) T : P3 → P2 , where T (p(x)) = p0 (x), the formal derivative.
Z 1
c) T : P1 → R, where T (p(x)) =
p(x) dx.
0
12. For each of the following, determine whether the linear operator T : Rn → Rn is one-to-one. If so,
find also the inverse linear operator T −1 : Rn → Rn :
a) T (x1 , x2 , x3 , . . . , xn ) = (x2 , x1 , x3 , . . . , xn )
b) T (x1 , x2 , x3 , . . . , xn ) = (x2 , x3 , . . . , xn , x1 )
c) T (x1 , x2 , x3 , . . . , xn ) = (x2 , x2 , x3 , . . . , xn )
13. Consider the operator T : R2 → R2 , where T (x1 , x2 ) = (x1 + kx2 , −x2 ) for every (x1 , x2 ) ∈ R2 .
Here k ∈ R is fixed.
a) Show that T is a linear operator.
b) Show that T is one-to-one.
c) Find the inverse linear operator T −1 : R2 → R2 .
14. Consider the linear transformation T : P2 → P1 , where T (p0 + p1 x + p2 x2 ) = (p0 + p2 ) + (2p0 + p1 )x
for every polynomial p0 + p1 x + p2 x2 in P2 .
a) Find the matrix for T with respect to the bases {1, x, x2 } and {1, x}.
b) Find T (2 + 3x + 4x2 ) by using the matrix A.
c) Use the matrix A to recover the formula T (p0 + p1 x + p2 x2 ) = (p0 + p2 ) + (2p0 + p1 )x.
15. Consider the linear operator T : R2 → R2 , where T (x1 , x2 ) = (x1 −x2 , x1 +x2 ) for every (x1 , x2 ) ∈ R2 .
a) Find the matrix A for T with respect to the basis {(1, 1), (−1, 0)} of R2 .
b) Use the matrix A to recover the formula T (x1 , x2 ) = (x1 − x2 , x1 + x2 ).
c) Is T one-to-one? If so, use the matrix A to find the inverse linear operator T −1 : R2 → R2 .
Chapter 8 : Linear Transformations
page 32 of 35
c
Linear Algebra
W W L Chen, 1997, 2008
16. Consider the real vector space of all real sequences x = (x1 , x2 , x3 , . . .) such that the series
∞
X
xn
n=1
is convergent.
a) Show that the transformation T : V → R, given by
T (x) =
∞
X
xn
n=1
for every x ∈ V , is a linear transformation.
b) Is the linear transformation T one-to-one? If so, give a proof. If not, find two distinct vectors
x, y ∈ V such that T (x) = T (y).
17. Suppose that T1 : R2 → R2 and T2 : R2 → R2 are linear operators such that
T1 (x1 , x2 ) = (x1 + x2 , x1 − x2 )
and
T2 (x1 , x2 ) = (2x1 + x2 , x1 − 2x2 )
for every (x1 , x2 ) ∈ R2 .
a) Show that T1 and T2 are one-to-one.
b) Find the formulas for T1−1 , T2−1 and (T2 ◦ T1 )−1 .
c) Verify that (T2 ◦ T1 )−1 = T1−1 ◦ T2−1 .
18. Consider the transformation T : P1 → R2 , where T (p(x)) = (p(0), p(1)) for every polynomial p(x)
in P1 .
a) Find T (1 − 2x).
b) Show that T is a linear transformation.
c) Show that T is one-to-one.
d) Find T −1 (2, 3), and sketch its graph.
19. Suppose that V and W are finite dimensional real vector spaces with dim V > dim W . Suppose
further that T : V → W is a linear transformation. Explain why T cannot be one-to-one.
20. Suppose that

1
A = 2
6
3
0
−2

−1
5 
4
is the matrix for a linear operator T : P2 → P2 with respect to the basis B = {p1 (x), p2 (x), p3 (x)}
of P2 , where p1 (x) = 3x + 3x2 , p2 (x) = −1 + 3x + 2x2 and p3 (x) = 3 + 7x + 2x2 .
a) Find [T (p1 (x))]B , [T (p2 (x))]B and [T (p3 (x))]B .
b) Find T (p1 (x)), T (p2 (x)) and T (p3 (x)).
c) Find a formula for T (p0 + p1 x + p2 x2 ).
d) Use the formula in part (c) to compute T (1 + x2 ).
21. Suppose that B = {v1 , v2 , v3 , v4 } is a basis for a real vector space V . Suppose that T : V → V is a
linear operator, with T (v1 ) = v2 , T (v2 ) = v4 , T (v3 ) = v1 and T (v4 ) = v3 .
a) Find the matrix for T with respect to the basis B.
b) Is T one-to-one? If so, describe its inverse.
Chapter 8 : Linear Transformations
page 33 of 35
c
Linear Algebra
W W L Chen, 1997, 2008
22. Let Pk denote the vector space of all polynomials with real coefficients and degree at most k.
Consider P2 with basis B = {1, x, x2 } and P3 with basis C = {1, x, x2 , x3 }. We define T1 : P2 → P3
and T2 : P3 → P2 as follows. For every polynomial p(x) = a0 + a1 x + a2 x2 in P2 , we have
T1 (p(x)) = xp(x) = a0 x + a1 x2 + a2 x3 . For every polynomial q(x) in P3 , we have T2 (q(x)) = q 0 (x),
the formal derivative of q(x) with respect to the variable x.
a) Show that T1 : P2 → P3 and T2 : P3 → P2 are linear transformations.
b) Find T1 (1), T1 (x), T1 (x2 ), and compute the matrix A1 for T1 : P2 → P3 with respect to the bases
B and C.
c) Find T2 (1), T2 (x), T2 (x2 ), T2 (x3 ), and compute the matrix A2 for T2 : P3 → P2 with respect to
the bases C and B.
d) Let T = T2 ◦ T1 . Find T (1), T (x), T (x2 ), and compute the matrix A for T : P2 → P2 with respect
to the basis B. Verify that A = A2 A1 .
23. Suppose that T : V → V is a linear operator on a real vector space V with basis B. Suppose that
for every v ∈ V , we have


 
x1 − x2 + x3
x1
[T (v)]B =  x1 + x2 
and
[v]B =  x2  .
x1 − x2
x3
a) Find the matrix for T with respect to the basis B.
b) Is T one-to-one? If so, describe its inverse.
24. For each of the following, let V be the subspace with basis B = {f1 (x), f2 (x), f3 (x)} of the space
of all real valued functions defined on R. Let T : V → V be defined by T (f (x)) = f 0 (x) for every
function f (x) in V . Find the matrix for T with respect to the basis B:
a) f1 (x) = 1, f2 (x) = sin x, f3 (x) = cos x
b) f1 (x) = e2x , f2 (x) = xe2x , f3 (x) = x2 e2x
25. Let P2 denote the vector space of all polynomials with real coefficients and degree at most 2,
with basis B = {1, x, x2 }. Consider the linear operator T : P2 → P2 , where for every polynomial
p(x) = a0 + a1 x + a2 x2 in P2 , we have T (p(x)) = p(2x + 1) = a0 + a1 (2x + 1) + a2 (2x + 1)2 .
a) Find T (1), T (x), T (x2 ), and compute the matrix A for T with respect to the basis B.
b) Use the matrix A to compute T (3 + x + 2x2 ).
c) Check your calculations in part (b) by computing T (3 + x + 2x2 ) directly.
d) What is the matrix for T ◦ T : P2 → P2 with respect to the basis B?
e) Consider a new basis B 0 = {1 + x, 1 + x2 , x + x2 } of P2 . Using a change of basis matrix, compute
the matrix for T with respect to the basis B 0 .
f) Check your answer in part (e) by computing the matrix directly.
26. Consider the linear operator T : P1 → P1 , where for every polynomial p(x) = p0 + p1 x in P1 , we
have T (p(x)) = p0 + p1 (x + 1). Consider also the bases B = {6 + 3x, 10 + 2x} and B 0 = {2, 3 + 2x}
of P1 .
a) Find the matrix for T with respect to the basis B.
b) Use Proposition 8V to compute the matrix for T with respect to the basis B 0 .
27. Suppose that V and W are finite dimensional real vector spaces. Suppose further that B and B 0 are
bases for V , and that C and C 0 are bases for W . Show that for any linear transformation T : V → W ,
we have
[I]C 0 ,C [T ]C,B [I]B,B0 = [T ]C 0 ,B0 .
28. Prove Proposition 8W.
29. Prove Proposition 8X.
Chapter 8 : Linear Transformations
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Linear Algebra
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30. For each of the following linear transformations T : R3 → R3 , find a basis B of R3 such that the
matrix for T with respect to the basis B is a diagonal matrix:
a) T (x1 , x2 , x3 ) = (−x2 + x3 , −x1 + x3 , x1 + x2 )
b) T (x1 , x2 , x3 ) = (4x1 + x3 , 2x1 + 3x2 + 2x3 , x1 + 4x3 )
31. Consider the linear operator T : P2 → P2 , where
T (p0 + p1 x + p2 x2 ) = (p0 − 6p1 + 12p2 ) + (13p1 − 30p2 )x + (9p1 − 20p2 )x2 .
a) Find the eigenvalues of T .
b) Find a basis B of P2 such that the matrix for T with respect to B is a diagonal matrix.
Chapter 8 : Linear Transformations
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