Transcript + T

Elementary Linear Algebra
Anton & Rorres, 9th Edition
Lecture Set – 08
Chapter 8:
Linear Transformations
Chapter Content






General Linear Transformations
Kernel and Range
Inverse Linear Transformations
Matrices of General Linear Transformations
Similarity
Isomorphism
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Elementary Linear Algebra
2
Linear Transformation

Definition
 If T : V  W is a function from a vector space V into a
vector space W, then T is called a linear transformation
from V to W if for all vectors u and v in V and all scalars c
 T (u + v) = T (u) + T (v)
 T (cu) = cT (u)
In the special case where V = W, the linear transformation T :
V  V is called a linear operator on V.
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Linear Transformation

Example (Zero Transformation)
 The mapping T : V  W such that T(v) = 0 for every v in V
is a linear transformation called the zero transformation.

Example (Identity Operator)
 The mapping I : V  I defined by I (v) = v is called the
identity operator on V.
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Orthogonal Projections



Suppose that W is a finite-dimensional subspace of an inner product
space V ; then the orthogonal projection of V onto W is the
transformation defined by
T (v) = projWv
If S = {w1, w2, …, wr} is any orthogonal basis for W, then T (v) is
given by the formula
T (v) = projWv = v, w1 w1 + v, w2 w2 + ··· + v, wr wr
This projection a linear transformation:
 T(u + v) = T(u) + T(v)
 T(cu) = cT(u)
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A Linear Transformation from a Space V
to Rn



Let S = {w1, w2, …, wn} be a basis for an n-dimensional vector space
V, and let
(v)s = (k1,, k2,, …, kn)
be the coordinate vector relative to S of a vector v in V; thus v =
k1w1 + k2w2 + …+ kn wn
Define T : V  Rn to be the function that maps v into its coordinate
vector relative to S; that is,
T (v) = (v)s = (k1,, k2,, …, kn)
Then the function T is a linear transformation:


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Let u = c1w1 + c2w2 + …+ cn wn and v = d1w1 + d2w2 + …+ dn wn
Check if (u + v)s = (u)s + (v)s and (ku)s = k(u)s
Elementary Linear Algebra
6
A Linear Transformation from Pn to Pn+1

Let p = p(x) = c0 + c1x + ··· + cnx n be a polynomial in Pn ,
and define the function T : Pn  Pn+1 by
T (p) = T (p(x)) = xp(x) = c0x + c1x2 + ··· + cnx n+1

The function T is a linear transformation:

For any scalar k and any polynomials p1 and p2 in Pn we have


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T (p1 + p2) = T (p1(x) + p2 (x)) = x (p1(x) + p2 (x)) = x p1(x) + x p2 (x)
= T (p1) + T (p2)
T (k p) = T (k p(x)) = x (k p(x)) = k (x p(x))= k T(p)
Elementary Linear Algebra
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A Linear Transformation Using an Inner
Product

Let V be an inner product space and let v0 be any fixed
vector in V.
Let T : V  R be the transformation that maps a vector v
into its inner product with v0; that is,
T (v) = v, v0

From the properties of an inner product



T (u + v) = u + v, v0 = u, v0 + v, v0
T (k u) = k u, v0 = k u, v0 = kT (u)
Thus, T is a linear transformation.
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8
Properties of Linear Transformation



If T : V  W is a linear transformation, then for any vectors v1 and
v2 in V and any scalars c1 and c2, we have
T (c1v1 + c2v2) = T (c1v1) + T (c2v2) = c1T (v1) + c2T (v2)
More generally, if v1 , v2 , …, vn are vectors in V and c1 , c2 , …, cn are
scalars, then
T (c1v1 + c2v2 +…+ cnvn ) = c1T (v1) + c2T (v2) +…+ cnT (vn)
The above equation is sometimes described by saying that linear
transformations preserve linear combinations.
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Theorem

Theorem 8.1
 If T : V  W is a linear transformation, then
 T(0) = 0
 T(-v) = -T(v) for all v in V
 T(v – w) = T(v) – T(w) for all v and w in V
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Finding Linear Transformations from
Images of Basis

If T : V  W is a linear transformation, and if {v1 , v2 , …, vn } is any
basis for V, then the image T (v) of any vector v in V can be
calculated from the images
T (v1), T (v2), …, T (vn)
of the basis vectors.

This can be done by first expressing v as a linear combination of the
basis vectors, say
v = c1 v1+ c2 v2+ …+ cn vn
and then the transformation becomes
T (v) = c1 T (v1) + c2 T (v2) + … + cn T (vn)

A linear transformation is completely determined by its images of
any basis vectors.
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Example

Consider the basis S = {v1 , v2 , v3} for R3 , where
v1 = (1,1,1), v2 = (1,1,0), and v3 = (1,0,0).
Let T: R3  R2 be the linear transformation such that
T (v1) = (1,0), T (v2) = (2,-1), T (v3) = (4,3).
Find a formula for T (x1, x2, x3); then use this formula to compute
T (2, -3, 5).
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Composition of T2 with T1

Definition


If T1 : U  V and T2 : V  W are linear transformations, the
composition of T2 with T1, denoted by T2  T1 (read “T2
circle T1 ”), is the function defined by the formula
(T2  T1 )(u) = T2 (T1 (u))
where u is a vector in U.
Theorem 8.1.2

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If T1 : U  V and T2 : V  W are linear transformations,
then (T2  T1 ) : U  W is also a linear transformation.
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Remark


The compositions can be defined for more than two linear
transformations.
For example, if T1 : U  V and T2 : V  W ,and T3 : W  Y
are linear transformations, then the composition T3  T2  T1
is defined by (T3  T2  T1 )(u) = T3 (T2 (T1 (u)))
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Chapter Content






General Linear Transformations
Kernel and Range
Inverse Linear Transformations
Matrices of General Linear Transformations
Similarity
Isomorphism
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Elementary Linear Algebra
16
Kernel and Range

Recall:
 If A is an mn matrix, then the nullspace of A consists of all
vector x in Rn such that Ax = 0.
 The column space of A consists of all vectors b in Rm for
which there is at least one vector x in Rn such that Ax = b.
 The nullspace of A consists of all vectors in Rn that
multiplication by A maps into 0. (in terms of matrix
transformation)
m
 The column space of A consists of all vectors in R that are
images of at least one vector in Rn under multiplication by A.
(in terms of matrix transformation)
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Kernel and Range

Definition


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If T : V  W is a linear transformation, then the set of
vectors in V that T maps into 0 is called the kernel of T; it is
denoted by ker(T).
The set of all vectors in W that are images under T of at
least one vector in V is called the range of T; it is denoted by
R(T).
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Examples



If TA : Rn  Rm is multiplication by the mn matrix A, then the
kernel of TA is the nullspace of A and the range of TA is the column
space of A.
Let T : V  W be the zero transformation. Since T maps every
vector in V into 0, it follows that ker(T) = V. Moreover, since 0 is the
only image under T of vectors in V, we have R(T) = {0}.
Let I : V  V be the identity operator. Since I (v) = v for all vectors
in V, every vector in V is the image of some vector; thus, R(I) = V.
Since the only vector that I maps into 0 is 0, it follows ker(I) = {0}.
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Example


Let T : R3  R3 be the orthogonal projection on the xy-plane. The
kernel of T is the set of points that T maps into 0 = (0,0,0); these are
the points on the z-axis.
Since T maps every points in R3 into the xy-plane, the range of T must
be some subset of this plane.
But every point (x0 ,y0 ,0) in the xy-plane is the image under T of
some point. Thus R(T) is the entire xy-plane.
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Example



Let T : R2  R2 be the linear operator that rotates each vector
in the xy-plane through the angle .
Since every vector in the xy-plane can be obtained by rotating
through some vector through angle , we have R(T) = R2.
The only vector that rotates into 0 is 0, so ker(T) = {0}.
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Properties of Kernel and Range

Theorem 8.2.1

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If T : V  W is linear transformation, then:
 The kernel of T is a subspace of V.
 The range of T is a subspace of W.
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Properties of Kernel and Range

Definition
 If T : V  W is a linear transformation, then the dimension
of the range of T is called the rank of T and is denoted by
rank(T).
 The dimension of the kernel is called the nullity of T and is
denoted by nullity(T).

Theorem 8.2.2
 If A is an mn matrix and TA : Rn  Rm is multiplication by
A, then:
 nullity (TA) = nullity (A)
 rank (TA) = rank (A)
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Example

1
0

0

0
Let TA : R6  R4 be multiplication by
5  3
 1 2 0 4
 3 7 2 0

1
4

A
 2 5 2 4
6
1


4

9
2

4

4
7


0 4 28 37 13
1 2 12 16 5 
0 0
0
0
0

0 0
0
0
0
 x1 
 4   28 37 
 13
x 
 2  12  16 
 5 
 2
     


 x3 
1   0   0 
 0 
   r   s t   u

 x4 
0  1   0 
 0 
 x5 
0  0   1 
 0 
 
     


 0   0   0 
 1 
 x6 
Find the rank and nullity of TA


In Example 1 of Section 5.6 we showed that rank (A) = 2 and nullity
(A) = 4. (use reduced row-echelon form, etc.)
Thus, from Theorem 8.2.2, rank (TA) = 2 and nullity (TA) = 4.
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Example



Let T : R3  R3 be the orthogonal projection on the xyplane.
From Example 4, the kernel of T is the z-axis, which is
one-dimensional; and the range of T is the xy-plane,
which is two-dimensional.
Thus, nullity (T) = 1 and rank (T) = 2.
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Dimension Theorem for Linear
Transformations

Theorem 8.2.3
 If T : V  W is a linear transformation from an ndimensional vector space V to a vector space W, then
rank(T) + nullity(T) = n

Remark
 In words, this theorem states that for linear transformations
the rank plus the nullity is equal to the dimension of the
domain.
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Chapter Content






General Linear Transformations
Kernel and Range
Inverse Linear Transformations
Matrices of General Linear Transformations
Similarity
Isomorphism
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Elementary Linear Algebra
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One-to-One Linear Transformation


A linear transformation T : V  W is said to be oneto-one if T maps distinct vectors in V into distinct
vectors in W.
Examples

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If A is an nn matrix and TA : Rn  Rn is multiplication
by A, then TA is one-to-one if and only if A is an
invertible matrix (Theorem 4.3.1).
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Example

Let T : R2  R2 be the linear operator that rotates
each vector in the xy-plane through an angle . We
showed that ker(T) = {0} and R(T) = R2.

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Thus, rank(T) + nullity(T) = 2 + 0 = 2.
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Theorem 8.3.1 (Equivalent Statements)

If T : V  W is a linear transformation, then the
following are equivalent.

T is one-to-one

The kernel of T contains only zero vector; that is, ker(T)
= {0}

Nullity(T) = 0
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Theorem 8.3.2

If V is a finite-dimensional vector space and T : V  V is
a linear operator, then the following are equivalent.

T is one-to-one

ker(T) = {0}

Nullity(T) = 0

The range of T is V; that is, R(T) = V
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Example

Let TA : R4  R4 be multiplication by
1
2
A
3

1
3  2 4
6  4 8
9 1 5

1 4 8
Determine whether TA is one to one.

Solution:
 det(A) = 0, since the first two rows of A are proportional
 A is not invertible
 TA is not one-to-one.
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Inverse Linear Transformations

If T : V  W is a linear transformation, then the range of T
denoted by R (T), is the subspace of W consisting of all images
under T of vectors in V.

If T is one-to-one, then each vector w in R(T) is the image of a
unique vector v in V.

This uniqueness allows us to define a new function, call the
inverse of T, denoted by T –1, which maps w back into v.

The mapping T –1 : R (T)  V is a linear transformation.
Moreover,
T –1(T (v)) = T –1(w) = v
T –1(T (w)) = T –1(v) = w
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Inverse Linear Transformations



If T : V  W is a one-to-one linear transformation,
then the domain of T –1 is the range of T.
The range may or may not be all of W (one-to-one but
not onto).
For the special case that T : V  V, then the linear
transformation is one-to-one and onto.
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Example (An Inverse Transformation)


Let T : R3  R3 be the linear operator defined by the formula
T (x1, x2, x3) = (3x1 + x2, -2x1 – 4x2 + 3x3, 5x1 + 4 x2 – 2x3).
Determine whether T is one-to-one; if so, find T -1(x1,x2,x3) .
Solution:
3 1 0
[T ]   2  4 3 
 5 4  2
 4  2  3
[T ]1    11 6 9 
 12 7 10 
  x1  
 x1   4  2  3  x1  4 x1  2 x2  3x3 


T 1   x2    [T 1 ]  x2     11 6 9   x2    11x1  6 x2  9 x3 
 x  
 x3   12 7 10   x3   12 x1  7 x2  10 x3 
 3
T 1 ( x1 , x2 , x3 )  (4 x1  2 x2  3x3 ,11x1  6 x2  9 x3 ,12 x1  7 x2  10 x3 )
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Theorem 8.3.3

If T1 : U  V and T2 : V  W are one to one linear
transformation then:


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T2  T1 is one to one
(T2  T1)-1 = T1-1  T2-1
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36
Chapter Content






General Linear Transformations
Kernel and Range
Inverse Linear Transformations
Matrices of General Linear Transformations
Similarity
Isomorphism
2016/4/8
Elementary Linear Algebra
37
Matrices of General Linear
Transformations

Remark:

If V and W are finite-dimensional vector spaces (not
necessarily Rn and Rm), then any transformation T : V  W
can be regarded as a matrix transformation.

The basic idea is to work with coordinate matrices of the
vectors rather than with the vectors themselves.
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Matrices of Linear Transformations

Suppose V and W are n and m dimensional vector space and B
and B are bases for V and W, then for x in V, the coordinate
matrix [x]B will be a vector in Rn, and coordinate matrix [T(x)]
m
B will be a vector in R .
T
A vector in V
(n-dimensional)
x
A vector in Rn
[x]B
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T (x)
?
[T (x)]B
Elementary Linear Algebra
A vector in W
(m-dimensional)
A vector in Rm
39
Matrices of Linear Transformations
T maps V into W

If we let A be the standard
matrix for this
transformation, then A [x]B =
[T (x)]B

The matrix A is called the
matrix for T with respect to
the bases B and B
T
T (x)
x
[x]B
A
[T (x)]B
Multiplication by A
maps Rn to Rm
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Matrices of Linear Transformations





Let B = {u1, …, un} be a basis for the n-dimensional space V and
B = {u1, …, um} be a basis for the m-dimensional space W.
 a11 a12  a1n 
Consider an mn matrix
a

a

a
21
22
2
n

A
 

 


a
a

a
m2
mn 
 m1
such that A [x]B = [T(x)]B holds for all vectors x in V.
That is, A [x]B = [T(x)]B has to hold for the basis vectors u1, …, un.
Thus, we need
A [u1]B = [T(u1)]B , A [u2]B = [T(u2)]B , …, A [un]B = [T(un)]B
Since
[u1]B = e1 , [u2]B = e2 , …, [un]B = en
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Matrices of Linear Transformations

We have

Thus, A  [[T (u1 )]B' | [T (u 2 )]B' |  | [T (u n )]B',] which is the matrix for T
w.r.t. the bases B and B, and denoted by the symbol [T]B,B
That is,

 a11 
 a1n 
a 
a 
[T (u1 )]B '  A[u1 ]B  A e1   21 , ...... , T [(u1 )]B '  A[u n ]B  A e n   2 n 
  
  
 
 
a
 m1 
amn 
[T ]B ', B  [[T (u1 )] B ' | [T (u 2 )] B ' |  | [T (u n )] B ' ]
and
[T ]B ', B [x]B  [T (x)] B '
Basis for the image space
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Basis for the domain
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42
Matrices for Linear Operators


In the special case where V = W, the resulting matrix is called
the matrix for T with respect to the basis B and denoted by [T]B
rather than [T]B,B.
If B = {u1, …, un} , then we have
[T ]B  [[T (u1 )]B | [T (u 2 )]B |  | [T (u n )]B ]
and

[T ]B [x]B  [T (x)]B
That is, the matrix for T times the coordinate matrix for x is
the coordinate matrix for T(x).
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43
Example
Let T : P1  P2 be the transformations defined by
T (p(x)) = xp(x).
Find the matrix for T with respect to the standard bases,
B = {u1, u2} and B = {v1, v2, v3},
where u1 = 1, u2 = x ; v1 = 1, v2 = x , v3 = x2
 Solution:
 T(u1) = T(1) = (x)(1) = x
and T(u2) = T(x) = (x)(x) = x2
T
 [T (u1)]B’ = [0 1 0]
[T (u2)]B’ = [0 0 1]T
 Thus, the matrix for T w.r.t. B and B’ is
0 0 
[T ]B ', B  [[T (u1 )]B ' | [T (u 2 )]B ' ]  1 0
0 1
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
Elementary Linear Algebra
Example


Let T : R2  R3 be the linear transformation defined by
x2


  x1   
T        5 x1  13x2 
  x2    7 x  16 x 
1
2

Find the matrix for the transformation T with respect to the
bases
B = {u1,u2} for R2 and B = {v1,v2,v3} for R3, where
1
 1
0 
3
5 
u1   , u 2   , v1   0 , v 2   2 , v 3  1
1
 2
 1
 2 
2
1
2
 Solution: T (u )   2  v  2 v , T (u )   1   3v  v  v
1
1
3
2
1
2
3
 
 
  5
 3
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45
Example
1
2
T (u1 )   2  v1  2 v 3 , T (u 2 )   1   3v1  v 2  v 3
  5
 3
1
3
[T (u1 )]B '   0 , [T (u 2 )] B '   1 
 2
 1
3
1
[T ]B ', B  [[T (u1 )]B ' | [T (u 2 )]B ' ]   0
1 
 2  1
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46
Theorems

Theorem 8.4.1


If T : Rn  Rm is a linear transformation and if B and B are
the standard bases for Rn and Rm, respectively, then
[T]B,B = [T]
Theorem 8.4.2

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If T1 : U  V and T2 : V  W are linear transformations,
and if B, B and B are bases for U, V and W, respectively,
then
[T2  T1]B,B’ = [T2 ]B’,B’’[T1 ]B’’,B
Elementary Linear Algebra
47
Theorem 8.4.3

If T : V  V is a linear operator and if B is a basis for V
then the following are equivalent



T is one to one
[T]B is invertible
Moreover, when these equivalent conditions hold
[T-1]B = [T]B-1
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48
Indirect Computation of a Linear
Transformation

An indirect procedure to compute a linear transformation:
1) Compute the coordinate matrix [x]B
2) Multiply [x]B on the left by [T]B,B to produce [T (x)]B
3) Reconstruct T (x) from its coordinate matrix [T (x)]B
x
Direction
computation
(3)
(1)
[x]B
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T (x)
Multiply by [T]B,B
(2)
Elementary Linear Algebra
[T (x)]B
49
Example


Let T : P2  P2 be linear operator defined by T(p(x)) = p(3x – 5),
that is, T (co + c1x + c2x2) = co + c1(3x – 5) + c2(3x – 5)2
2
 Find [T]B with respect to the basis B = {1, x, x }
 Use the indirect procedure to compute T (1 + 2x + 3x2)
 Check the result by computing T (1 + 2x + 3x2)
Solution:
 Form the formula for T,
T(1) = 1, T(x) = 3x – 5, T(x2) = (3x – 5)2 = 9x2 – 30x + 25
 Thus,
1  5 25 
[T ]B  0 3  30
0 0
9 
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50
Example


1  5 25 
[T ]B  0 3  30
0 0
9 
The coordinate matrix relative to B for vector p = 1 + 2x + 3x2 is
[p]B = [1 2 3]T.
1  5 25  1  66 

  

Thus, [T (1 + 2x + 3x2)]B = [T (p)]B = [T]B [p]B = 0 3  30 2   84
0 0
9  3  27 
 T (1 + 2x + 3x2) = 66 – 84x + 27x2

By direction computation:

T (1 + 2x + 3x2) = 1 + 2(3x – 5) + 3(3x – 5)2
= 1 + 6x – 10 + 27x2 – 90x + 75
= 66 – 84x + 27x2
x
Direction
computation
(3)
(1)
Multiply by [T]B,B
[x]B
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T (x)
(2)
[T (x)]B
51
Chapter Content






General Linear Transformations
Kernel and Range
Inverse Linear Transformations
Matrices of General Linear Transformations
Similarity
Isomorphism
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52
Similarity

The matrix of a linear operator T : V  V depends on the
basis selected for V that makes the matrix for T as simple
as possible – a diagonal or triangular matrix.
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53
Simple Matrices for Linear Operators




Consider the linear operator T : R2  R2 defined by T   x1     x1  x2 
  x    2 x  4 x 
2
1
2
 2 
and the standard basis B = {e1, e2} for R .
The matrix for T with respect to this basis is the standard matrix
for T;
that is, [T]B = [T] = [T(e1) | T(e2)].
Since T (e1) = [1 -2]T, T (e2) = [1 4]T, we have [T ]B   1 1
  2 4
However, if u1 = [1 1]T, u2 = [1 2]T, then the matrix for T with
respect to the basis B = {u1, u2} is the diagonal matrix
 2 0
[T ]B '  

0
3


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54
Theorem 8.5.1

If B and B are bases for a finite-dimensional vector
space V, and if I : V  V is the identity operator, then
[I]B,B is the transition matrix from B to B.

Remark
I
V
v
Basis = B
V
v
Basis = B
[I]B,B is the transition matrix from B to B.
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55
Theorem

Theorem 8.5.2
 Let T : V  V be a linear operator on a finite-dimensional
vector space V, and let B and B be bases for V. Then
[T]B = P-1 [T]B P
where P is the transition matrix from B to B.

Remark:
I
V
v
Basis = B
T
V v
Basis = B
V
I
T(v)
Basis = B
V
T(v)
Basis = B
[T]B = [I]B,B[T]B[I]B,B = P-1 [T]B P
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56
Example

Let T : R2  R2 be defined by
x   x  x 
T   1     1 2 
  x2    2 x1  4 x2 
Find the matrix T with respect to the standard basis B = {e1, e2}
for R2, then use Theorem 8.5.2 to find the matrix of T with
respect to the basis B = {u1, u2}, where u1 = [1 1]T and u2 =
[1 2]T.

 1 1
[T ]B  

  2 4
Solution:
P  [ I ]B.B '  [[u1 ' ]B | [u 2 ' ]B ]
 2  1
P 1  

 1 1 
 2  1  1 1 1 1 2 0
 P 1 T B P  







 1 1   2 4 1 2 0 3
1 1
P

1 2
T B '
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57
Definitions

Definition


If A and B are square matrices, we say that B is similar to A
if there is an invertible matrix P such that B = P-1AP
Definition

2016/4/8
A property of square matrices is said to be a similarity
invariant or invariant under similarity if that property is
shared by any two similar matrices.
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58
Similarity Invariants
Property
Description
Determinant
A and P-1AP have the same determinant.
Invertibility
A is invertible if and only if P-1AP is invertible.
Rank
A and P-1AP have the same rank.
Nullity
A and P-1AP have the same nullity.
Trace
A and P-1AP have the same trace.
Characteristic polynomial
A and P-1AP have the same characteristic polynomial.
Eigenvalues
A and P-1AP have the same eigenvalues
Eigenspace dimension
If  is an eigenvalue of A and P-1AP then the
eigenspace of A corresponding to  and the
eigenspace of P-1AP corresponding to  have the
same dimension.
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59
Determinant of A Linear Operator




Two matrices representing the same linear operator T : V  V with
respect to different bases are similar.
For any two bases B and B we must have
det([T]B) = det([T]B)
Thus we define the determinant of the linear operator T to be
det(T) = det([T]B)
where B is any basis for V.
Example
  x1    x1  x2 
2
2
 Let T : R  R be defined by
T      

x

2
x

4
x
1
2
 2 
 1 1
[T ]B  

  2 4
2 0
T B '   
0 3
2016/4/8
det(T )  6
det(T )  6
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60
Eigenvalues of a Linear Operator


A scalar  is called an eigenvalue of a linear operator T : V 
V if there is a nonzero vector x in V such that Tx = x. The
vector x is called an eigenvector of T corresponding to .
Equivalently, the eigenvectors of T corresponding to  are the
nonzero vectors in the kernel of I – T. This kernel is called
the eigenspace of T corresponding to .
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61
Eigenvalues of a Linear Operator

If V is a finite-dimensional vector space, and B is any basis
for V, then

The eigenvalues of T are the same as the eigenvalues of [T]B .

A vector x is an eigenvector of T corresponding to [T]B if and
only if its coordinate matrix [x]B is an eigenvector of [T]B
corresponding to .
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62
Example


Find the eigenvalues and bases for the eigenvalues of the
linear operator T : P2  P2 defined by
T (a + bx + cx2) = -2c + (a + 2b + c)x + (a + 3c)x2
Solution:
0 0  2 

1 
 The eigenvalues of T are  = 1 and  = 2 T B  1 2
1 0 3 
 The eigenvectors of [T]B are:
  2
 1
0
  = 2:
 = 1:
u1   0 , u 2  1
 1 
0
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Elementary Linear Algebra
u 3   1 
 1 
63
Example

Let T : R3  R3 be the linear operator given by
  x1     2 x3 


T   x2     x1  2 x2  x3 
  x    x  3x 
3

 3  1

Find a basis for R3 relative to which the matrix for T is
diagonal.
Solution:
 det(I  A)  3  52  8  4  (2)(2)(1)
 2 0 0
[T ]B '  0 2 0
0 0 1
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64
Onto Transformations




Let V and W be real vector spaces. We say that the linear
transformation T : V  W is onto if the range of T is W.
An onto transformation is also said to be surjective or to be a
surjection. For a surjective mapping, the range and the codomain
coincide.
If a transformation T : V  W is both one-to-one (also called
injective or an injection) and onto, then it is a one-to-one
mapping to its range W and so has an inverse T-1 : W  V.
A transformation that is one-to-one and onto is also said to be
bijective or to be a bijection between V and W.
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65
Theorem 8.6.1

Bijective Linear Transformation

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Let V and W be finite-dimensional vector spaces. If dim(V)
 dim(W), then there can be no bijective linear
transformation from V to W.
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66
Chapter Content






General Linear Transformations
Kernel and Range
Inverse Linear Transformations
Matrices of General Linear Transformations
Similarity
Isomorphism
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67
Isomorphisms

Definition
 An isomorphism between V and W is a bijective linear
transformation from V to W.
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68
Isomorphisms

Theorem 8.6.2 (Isomorphism Theorem)


Let V be a finite-dimensional real vector space. If dim(V) =
n, then there is an isomorphism from V to Rn.
Example

2016/4/8
The vector space P3 is isomorphic to R4, because the
transformation
T(a + bx + cx2 + dx3) = (a,b,c,d)
is one-to-one, onto, and linear.
Elementary Linear Algebra
69
Isomorphisms between Vector Spaces

Theorem 8.6.3 (Isomorphism of Finite-Dimensional Vector
Spaces)
 Let V and W be finite-dimensional vector spaces. If dim(V)
= dim(W), then V and W are isomorphic.
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70
Example

An Isomorphism between P3 and M22
 Because dim(P3) = 4 and dim(M22) = 4, these spaces are
isomorphic.
 We can find an isomorphism T : P3  M22:
1 0
0 1 
0 0
0 0
2
3
T (1)  
T ( x)  
T (x )  
T (x )  




0
0
0
0
1
0
0
1









2016/4/8
This is one-to-one and onto linear transformation, so it is an
isomorphism between P3 and M22.
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71