程序代写代做代考 C go 9.3

9.3
Week 10
Ali Mousavidehshikh
Department of Mathematics University of Toronto
Ali Mousavidehshikh
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Outline
1 9.3
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Given a linear transformation T : V → V , we would like to find a basis that makes the matrix MB(T) as simple as possible.
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9.3
Given a linear transformation T : V → V , we would like to find a basis that makes the matrix MB(T) as simple as possible.
IfU isasubspaceofV andT :V →V isalinear transformation, we define the image of T as T(U)={T(x):x ∈U}.
Definition: Let T : V → V be an operator. A subspace U is called T-invariant if T(U) ⊆ U. That is, for each x ∈ U we have T(x) ∈ U. So T : U → U is a linear operator.
Example: Let T : V → V be any linear operator.
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Example: Define T : R3 → R3 via T(a,b,c)=(3a+2b,b−c,4a+2b−c). Then
U = {(a,b,a) : a,b ∈ R} is a T-invariant subspace.
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Example: Define T : R3 → R3 via T(a,b,c)=(3a+2b,b−c,4a+2b−c). Then
U = {(a,b,a) : a,b ∈ R} is a T-invariant subspace. Solution Given (a, b, a) ∈ U, then
T(a,b,a) = (3a + 2b,b − a,3a + 2b) ∈ U.
Exercise: Let T : V → V be a linear operator, and suppose U = span{u1,…,uk} is a subspace of V. Show that U is T-invariant if and only if T(ui) ∈ U for each i = 1,…,k.
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Proof . Assume U is a T -invariant subspace with U ̸= {0} and U ̸= R2.
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Proof . Assume U is a T -invariant subspace with U ̸= {0} and U ̸= R2. Then dimU = 1,
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Proof . Assume U is a T -invariant subspace with U ̸= {0}
and U ̸= R2. Then dimU = 1, so U = Rx for some non-zero x ∈ R2.
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Proof . Assume U is a T -invariant subspace with U ̸= {0} and U ̸= R2. Then dimU = 1, so U = Rx for some non-zero x ∈ R2. Since U is T-invariant, we have T(x) ∈ U, say T(x) = rx for some x ∈ R.
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Proof . Assume U is a T -invariant subspace with U ̸= {0} and U ̸= R2. Then dimU = 1, so U = Rx for some non-zero x ∈ R2. Since U is T-invariant, we have T(x) ∈ U, say T(x) = rx for some x ∈ R. Let x = (a,b). This says that (b,−a) = T(a,b) = r(a,b) = (ra,rb). This gives b = ra and −a = rb. So −a = rb = r2a ⇒ (r2 + 1)a = 0. Hence,
a = 0. Then, b = ra = 0 too, a contradiction. Hence, no one-dimensional T-invariant subspace exists.
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Theorem: Let T : V → V be a linear operator, where V has dimension n, and suppose that U is any T-invariant subspace of V. Let B1 = {b1,…,bk} be any basis of U and extend it to a basis B = {b1,…,bk,bk+1,…,bn} of V in any way. Then MB(T) has block triangular form
􏰑MB1(T) Y􏰒
MB(T)= 0 Z ,whereZ is(n−k)×(n−k)and
MB1(T) is the matrix of the restriction of T to U.
Proof . Notice that
MB(T) = [CB(T(b1)) ··· CB(T(bn))]. The matrix
T : U → U (the restriction of T to U) with respect to the basis B1 is the k × k matrix
MB1(T) = [CB1(T(b1)) ··· CB1(T(bk))].
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Consequently, CB [T (b1)] = tk  = 0 0
in block
 .  .
0
form. A similar argument holds for CB(T(bi)) for
i = 2,3,…,k.
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t1 t2
 . 
. 􏰑CB1(T(b1))􏰒
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Consequently, CB [T (b1)] = tk  = 0 0
in block
 .  .
0
form. A similar argument holds for CB(T(bi)) for
i = 2,3,…,k.
t1 t2
 . 
. 􏰑CB1(T(b1))􏰒
Theorem: Let A be a block upper (lower) triangular matrix, A11 A12 A13 · · · A1n
0 A22 A23 ··· A2n 00A···A
say A =  33 3n, where the diagonal  . . . . 

0 0 0 ··· Ann blocks are square matrices. Then:
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(1) detA = detA11 detA22 ···detAnn.
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(1) detA = detA11 detA22 ···detAnn. (2) CA(x) = CA11(x)CA22(x)···CAnn(x).
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(1) detA = detA11 detA22 ···detAnn. (2) CA(x) = CA11(x)CA22(x)···CAnn(x).
Consider the linear operator T : P2 → P2 given by T(a+bx+cx2) = (−2a−b+2c)+(a+b)x+(−6a−2b+5c)x2. Show that U = span{x , 1 + 2x 2 } is T -invariant, use it to find a block upper triangular matrix for T, and use that to compute CT(x).
Solution. Since T(x) = x − (1 + 2×2) ∈ U and T(1+2×2)=x+2(1+2×2)∈U,wehaveU is
T -invariant.
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􏰑1 1􏰒 Notice that MB1(T) = −1 2 ,
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􏰑1 1􏰒 Notice that MB1(T) = −1 2 , so
􏰑MB1(T) Y􏰒
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􏰑 0 􏰒
MB(T)= 0 Z ,whereY = −2 andZ =(1).
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􏰑1 1􏰒 Notice that MB1(T) = −1 2 , so
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􏰑 0 􏰒
MB(T)= 0 Z ,whereY = −2 andZ =(1).In
􏰑MB1(T) Y􏰒
particular, CT(x) = CMB1(T)(x)CZ(x) = (x2 −3x +3)(x −1).
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􏰑1 1􏰒 Notice that MB1(T) = −1 2 , so
􏰑 0 􏰒
MB(T)= 0 Z ,whereY = −2 andZ =(1).In
􏰑MB1(T) Y􏰒
particular, CT(x) = CMB1(T)(x)CZ(x) = (x2 −3x +3)(x −1).
A real number λ is called an eigenvalue of a linear operator T : V → V if T(v) = λv holds for some non-zero vector
v ∈ V . In this case, v is called an eigenvector of T corresponding to λ, Eλ(T) = {v ∈ V : T(v) = λv} is called the eigenspace of T corresponding to λ.
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Theorem: Let T : V → V be a linear operator where
dimV = n, and let B denote any ordered basis of V, and let CB : V → Rn denote the coordinate isomorphism. Then:
(1) The eigenvalues λ of T are precisely the eigenvalues of the matrix MB(T) and thus are the roots of the characteristic polynomial CT(x).
(2) In this case the eigenspace Eλ(T) and Eλ(MB(T)) are isomorphic via the restriction CB : Eλ(T) → Eλ(MB(T)).
The previous theorem tells us that v ∈ Eλ(T) if and only if CB(v) ∈ Eλ(MB(T)). Notice that even though T and MB(T) have the same eigenvalues, the eigenvectors corresponding to
these differ (one must apply CB or C−1 to these eigenvectors B
(depending on direction)).
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Example: Find the eigenvalues and eigenspace for
T : P2 → P2 given by
T(a+bx +cx2) = (2a+b+c)+(2a+b−2c)x −(a+2c)x2.
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9.3
Example: Find the eigenvalues and eigenspace for
T : P2 → P2 given by
T(a+bx +cx2) = (2a+b+c)+(2a+b−2c)x −(a+2c)x2. Solution. Let B be the standard basis for P2.
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9.3
Example: Find the eigenvalues and eigenspace for
T : P2 → P2 given by
T(a+bx +cx2) = (2a+b+c)+(2a+b−2c)x −(a+2c)x2. Solution. Let B be the standard basis for P2. Then
211
MB(T) =  2 1 −2. Hence,
−1 0 −2
CT (x) = (x + 1)2(x − 3). Moreover,
   −1 
E−1(MB(T)) = span  2  and 1
  5
E3(MB(T)) = span  6  ,  −1 
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Notice that −1 + 2x + x2 is an eigenvector of T corresponding to
the eigenvalue −1 and CB(−1 + 2x + x2) is an eigenvector of
MB(T) corresponding to the eigenvalue −1 (CB and C−1 allow B
you to go back and forth between Eλ(T) and Eλ(MB(T)).
Each eigenspace of a linear operator T : V → V is a T-invariant subspace of V.
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Let U ={A∈Mnn :AT =A} and W = {B ∈ Mnn : BT = −B}.
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Let U ={A∈Mnn :AT =A} and
W={B∈Mnn:BT =−B}.GivenamatrixC∈Mnn,we
have C = 1(A+AT)+ 1(A−AT). 22
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Let U ={A∈Mnn :AT =A} and
W={B∈Mnn:BT =−B}.GivenamatrixC∈Mnn,we
have C = 1(A+AT)+ 1(A−AT). Letting X = 1(A+AT) 222
and Y = 1 (A − AT ), we have X ∈ U and Y ∈ W . 2
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9.3
Let U ={A∈Mnn :AT =A} and
W={B∈Mnn:BT =−B}.GivenamatrixC∈Mnn,we
have C = 1(A+AT)+ 1(A−AT). Letting X = 1(A+AT) 222
and Y = 1 (A − AT ), we have X ∈ U and Y ∈ W . In 2
particular, every matrix in Mnn can be written as the sum of two elements, one in U and the other in W . Moreover, this representation is unique: If A = P + Q where P ∈ U and
Q ∈ W , then
AT =PT +QT =P−Q ⇒ A+AT =2P andA−AT =2Q ⇒ P=1(A+AT)and
2
Q = 1(A−AT). 2
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Definition: If U and W are subspaces of a vector space V withU+W =V andU∩W ={0},thenwesayV isa direct sum of U and W and write V = U ⊕W. The subspace W is called the complement of U in V (and vice versa).
Examples: (1) R5 = U ⊕ W , where
U = span{(a,0,0,0,0) : a ∈ R} and
W = span{(0,b,c,d,e) : b,c,d,e ∈ R}. (2)ForanysubspaceU ofRn,Rn =U⊕U⊥.
Theorem: Let U and W be subspaces of a finite dimension vector space V . The following are equivalent.
(1) V = U ⊕ W .
(2) Each vector v ∈ V can be written uniquely in the form v = u + w , where u ∈ U and w ∈ W .
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Theorem: If V is finite dimensional and V = U ⊕ W , then dimV =dimU+dimW.
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Theorem: If V is finite dimensional and V = U ⊕ W , then dimV =dimU+dimW.
Theorem: Let T : V → V be a linear operator where V is finite dimensional. Suppose V = U1 ⊕ U2 where U1 and U2 are both T-invariant. If B1 = {b1,…,bk} and
B2 = {bk+1, . . . , bn} are bases of U1 and U2, respectively, then B = {b1,…,bk,bk+1,…,bn} is a basis of V and
􏰑MB1(T) 0 􏰒
MB(T) = 0 MB2(T) , where MB1(T) and MB2(T)
are the matrices of the restriction of T to U1 and U2, respectively.
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Example: Let T : V → V be a linear operator such that T 2 = 1v . Define U1 = {v : T (v ) = v } and
U2 ={v ∈V :T(v)=−v}. Then,
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Example: Let T : V → V be a linear operator such that T 2 = 1v . Define U1 = {v : T (v ) = v } and
U2 ={v ∈V :T(v)=−v}. Then,
(a) U1 and U2 are subspaces of V .
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Example: Let T : V → V be a linear operator such that T 2 = 1v . Define U1 = {v : T (v ) = v } and
U2 ={v ∈V :T(v)=−v}. Then,
(a) U1 and U2 are subspaces of V .
(b) V = U1 ⊕ U2.
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