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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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
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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)
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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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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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Kernel and Range
Recall:
If A is an mn 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 mn 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 mn 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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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 nn 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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Chapter Content
General Linear Transformations
Kernel and Range
Inverse Linear Transformations
Matrices of General Linear Transformations
Similarity
Isomorphism
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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 mn 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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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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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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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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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
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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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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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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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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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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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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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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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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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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
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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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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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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
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det(T ) 6
det(T ) 6
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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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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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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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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 52 8 4 (2)(2)(1)
2 0 0
[T ]B ' 0 2 0
0 0 1
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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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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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Chapter Content
General Linear Transformations
Kernel and Range
Inverse Linear Transformations
Matrices of General Linear Transformations
Similarity
Isomorphism
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Isomorphisms
Definition
An isomorphism between V and W is a bijective linear
transformation from V to W.
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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
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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.
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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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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
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This is one-to-one and onto linear transformation, so it is an
isomorphism between P3 and M22.
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