linear algebra ii lecture 14 - ualberta.caxichen/math22514f/20141119_printable.pdf · linear...
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Characteristic Polynomial, Eigenvalue, Eigenvector and Diagonalization
Linear Algebra II Lecture 14
Xi Chen 1
1University of Alberta
November 19, 2014
Xi Chen Linear Algebra II Lecture 14
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Characteristic Polynomial, Eigenvalue, Eigenvector and Diagonalization
Outline
1 Characteristic Polynomial, Eigenvalue, Eigenvector andDiagonalization
Xi Chen Linear Algebra II Lecture 14
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Characteristic Polynomial, Eigenvalue, Eigenvector and Diagonalization
Eigenvalue, Eigenvector and Diagonalization
Let T : V → V be a linear endomorphism. We call λ aneigenvalue of T and v 6= 0 an eigenvector of Tcorresponding to λ if T (v) = λv. All eigenvectors v of Tcorresponding to λ lie in the eigenspace
{v : T (v) = λv} = {v : (λI − T )v} = K (λI − T )
of T corresponding to λ.If dim V <∞, we call
det(xI − [T ]B←B) = xn + a1xn−1 + ...+ an
the characteristic polynomial of T , which is independent ofthe choice of B.T (v) = λv⇒ f (T )(v) = f (λ)v for all polynomials f (x).
(Cayley-Hamilton) f (T ) = 0 for the characteristicpolynomial f (x) of T .
Xi Chen Linear Algebra II Lecture 14
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Characteristic Polynomial, Eigenvalue, Eigenvector and Diagonalization
Characteristic Polynomial, Eigenvalue, Eigenvectorand Diagonalization
We call T : V → V diagonalizable if there exists anordered basis B of V such that [T ]B←B is diagonal.Let V be a vector space of dimension n. Then a lineartransformation T : V → V is diagonalizable if and only if Thas n linearly independent eigenvectors v1,v2, ...,vn. Inaddition,
[T ]B←B =
λ1
λ2. . .
λn
for B = {v1,v2, ...,vn} if v1,v2, ...,vn are linearlyindependent eigenvectors of T corresponding toλ1, λ2, ..., λn.
Xi Chen Linear Algebra II Lecture 14
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Characteristic Polynomial, Eigenvalue, Eigenvector and Diagonalization
Eigenvalues, Eigenvectors and Diagonalization
Let λ1, λ2, ..., λm be all the distinct eigenvalues ofT : V → V . Then T is diagonalizable if and only if
V = K (λ1I − T ) + K (λ2I − T ) + ...+ K (λmI − T )
or equivalently,
dim V = dim K (λ1I−T )+dim K (λ2I−T )+...+dim K (λmI−T )
or equivalently,
dim V ≤ dim K (λ1I−T )+dim K (λ2I−T )+...+dim K (λmI−T )
T is diagonalizable⇒ f (T ) is diagonalizable for allpolynomials f (x).
Xi Chen Linear Algebra II Lecture 14
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Characteristic Polynomial, Eigenvalue, Eigenvector and Diagonalization
Examples of Eigenvalues, Eigenvectors andDiagonalization
Let T : R3 → R3 be a linear transformation with twoeigenvalues 1 and 2. If
dim K (I − T ) = 1 and dim K (2I − T ) = 1,
is T diagonalizable?No since dim K (I − T ) + dim K (2I − T ) < 3. There exists anordered basis B such that
[T ]B←B =
1 1 00 1 00 0 2
or
1 0 00 2 10 0 2
.
Xi Chen Linear Algebra II Lecture 14
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Characteristic Polynomial, Eigenvalue, Eigenvector and Diagonalization
Examples of Eigenvalues, Eigenvectors andDiagonalization
Let T : R3 → R3 be a linear transformation with twoeigenvalues 1 and 2. If
dim K (I − T ) = 1 and dim K (2I − T ) = 2,
is T diagonalizable?Yes since dim K (I − T ) + dim K (2I − T ) = 3. LetB = {v1,v2,v3} with {v1} a basis of K (I − T ) and {v2,v3} abasis of K (2I − T ). Then
[T ]B←B =
12
2
.
Xi Chen Linear Algebra II Lecture 14
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Characteristic Polynomial, Eigenvalue, Eigenvector and Diagonalization
Examples of Eigenvalues, Eigenvectors andDiagonalization
Let T : R3 → R3 be the linear transformation
T (x , y , z) = (x + y , y + z,2z)
Find the characteristic polynomial, eigenvalues andeigenvectors of T and find a basis B such that [T ]B←B isdiagonal if such B exists.Solution. Let C be the standard basis. Then
[T ]C←C =
1 1 00 1 10 0 2
.So the characteristic polynomial of T is
det(xI − [T ]C←C) = (x − 1)2(x − 2).
Xi Chen Linear Algebra II Lecture 14
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Characteristic Polynomial, Eigenvalue, Eigenvector and Diagonalization
Examples of Eigenvalues, Eigenvectors andDiagonalization
Solution (CONT). T has eigenvalues 1 and 2 with eigenvectors
{v : T (v) = v} = K (I − T )
=
x
yz
:
0 −1 00 0 −10 0 −1
xyz
=
000
= Span
1
00
{v : T (v) = 2v} = K (2I − T )
=
x
yz
:
1 −1 00 1 −10 0 0
xyz
=
000
= Span
1
11
T is not diagonalizable since Span{(1,0,0), (1,1,1)} 6= R3.
Xi Chen Linear Algebra II Lecture 14
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Characteristic Polynomial, Eigenvalue, Eigenvector and Diagonalization
Examples of Eigenvalues, Eigenvectors andDiagonalization
Let T : R3 → R3 be the linear transformation
T (x , y , z) = (x , y + z,2z)
Find the characteristic polynomial, eigenvalues andeigenvectors of T and find a basis B such that [T ]B←B isdiagonal if such B exists.Solution. Let C be the standard basis. Then
[T ]C←C =
1 0 00 1 10 0 2
.So the characteristic polynomial of T is
det(xI − [T ]C←C) = (x − 1)2(x − 2).
Xi Chen Linear Algebra II Lecture 14
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Characteristic Polynomial, Eigenvalue, Eigenvector and Diagonalization
Examples of Eigenvalues, Eigenvectors andDiagonalization
Solution (CONT). T has eigenvalues 1 and 2 with eigenvectors
{v : T (v) = v} = K (I − T )
=
x
yz
:
0 0 00 0 −10 0 −1
xyz
=
000
= Span
1
00
,0
10
{v : T (v) = 2v} = K (2I − T )
=
x
yz
:
1 0 00 1 −10 0 0
xyz
=
000
= Span
0
11
T is diagonalizable since
Span{(1,0,0), (0,1,0), (0,1,1)} = R3.
Xi Chen Linear Algebra II Lecture 14
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Characteristic Polynomial, Eigenvalue, Eigenvector and Diagonalization
Examples of Eigenvalues, Eigenvectors andDiagonalization
Solution (CONT). Let B = {(1,0,0), (0,1,0), (0,1,1)}. Then
[T ]B←B =
11
2
= P−1C←B[T ]C←CPC←B
=
1 0 00 1 10 0 1
−1 1 0 00 1 10 0 2
1 0 00 1 10 0 1
1 1 0
0 1 10 0 2
and
1 0 00 1 10 0 2
have the same characteristic polynomials and but they are notsimilar.
Xi Chen Linear Algebra II Lecture 14
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Characteristic Polynomial, Eigenvalue, Eigenvector and Diagonalization
Examples of Eigenvalues, Eigenvectors andDiagonalization
Let V be a vector space of finite dimension and T : V → V be alinear transformation satisfying T 2 = I. Show that T isdiagonalizable.
Proof.
Since T 2 = I, I − T 2 = (I − T )(I + T ) = 0⇒
R(I − T ) ⊂ K (I + T )⇒ rank(I − T ) ≤ dim K (I + T ).
By Rank Theorem, rank(I − T ) = dim V − dim K (I − T ).Therefore, dim K (I + T ) + dim K (I − T ) ≥ dim V
⇒ dim K (I + T ) + dim K (I − T ) = dim V
So T is diagonalizable.
Xi Chen Linear Algebra II Lecture 14
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Characteristic Polynomial, Eigenvalue, Eigenvector and Diagonalization
Examples of Eigenvalues, Eigenvectors andDiagonalization
Let T : M2×2(R)→ M2×2(R) be the linear transformation
T (A) =[3 14 3
]︸ ︷︷ ︸
D
A
Is T diagonalizable? If it is, find a basis B such that [T ]B←B isdiagonal.Solution. Let λ be an eigenvalue of T with correspondingeigenvector A. Then
T (A) = λA⇔ DA = λA⇔ (λI − D)A = 0
⇔ (λI − D)[v1 v2
]= 0
⇔ (λI − D)v1 = (λI − D)v2 = 0⇔ v1,v2 ∈ Nul(λI − D).
Xi Chen Linear Algebra II Lecture 14
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Characteristic Polynomial, Eigenvalue, Eigenvector and Diagonalization
Examples of Eigenvalues, Eigenvectors andDiagonalization
Solution (CONT). It suffices to find the eigenvalues andeigenspaces of D: D has characteristic polynomial x2 − 6x + 5and eigenvalues 1 and 5 with eigenspaces
Nul(I − D) = Span{[
1−2
]}and Nul(5I − D) = Span
{[12
]}.
Correspondingly, T has eigenvalues 1 and 5 with eigenspaces
K (I − T ) = Span{[
1 0−2 0
],
[0 10 −2
]}K (5I − T ) = Span
{[1 02 0
],
[0 10 2
]}.
Xi Chen Linear Algebra II Lecture 14
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Characteristic Polynomial, Eigenvalue, Eigenvector and Diagonalization
Examples of Eigenvalues, Eigenvectors andDiagonalization
Solution (CONT). So T diagonalizable and
[T ]B←B =
1
15
5
where
B =
{[1 0−2 0
],
[0 10 −2
],
[1 02 0
],
[0 10 2
]}.
We also see that the characteristic polynomial of T is(x − 1)2(x − 5)2.
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Characteristic Polynomial, Eigenvalue, Eigenvector and Diagonalization
Examples of Eigenvalues, Eigenvectors andDiagonalization
Let T : Pn → Pn be the linear transformation
T (f (x)) = f (x + 1)
Is T diagonalizable? If it is, find a basis B such that [T ]B←B isdiagonal.Solution. Let D = {1, x , ..., xn} be the standard basis. Then
[T ]D←D︸ ︷︷ ︸A
=[(j−1
i−1
)](n+1)×(n+1)
=
1 1 ... 1
1 .... n. . .
...1
.Then the characteristic polynomial of T or equivalently A is(x − 1)n and A 6= I.
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Characteristic Polynomial, Eigenvalue, Eigenvector and Diagonalization
Examples of Eigenvalues, Eigenvectors andDiagonalization
Solution (CONT). T is diagonalizable if and only if A is.Suppose that A is diagonalizable. Then there exists aninvertible matrix P such that PAP−1 is diagonal. And since thecharacteristic polynomial of A is (x − 1)n,
PAP−1 =
1
1. . .
1
= I ⇒ A = P−1IP = I.
Contradiction. So T is not diagonalizable.
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Characteristic Polynomial, Eigenvalue, Eigenvector and Diagonalization
Examples of Eigenvalues, Eigenvectors andDiagonalization
Let T : Pn → Pn be the linear transformation
T (f (x)) = f (1− x)
Is T diagonalizable? If it is, find a basis B such that [T ]B←B isdiagonal.Solution. Note that T 2 = I. Such T is always diagonalizable.The eigenvectors of T are (1
2 − x)k with correspondingeigenvalue (−1)k . So
[T ]B←B =
1−1
. . .(−1)n
(n+1)×(n+1)
where B = {1, 12 − x , (1
2 − x)2, ..., (12 − x)n}.
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Characteristic Polynomial, Eigenvalue, Eigenvector and Diagonalization
Examples of Eigenvalues, Eigenvectors andDiagonalization
Let T : Pn → Pn be the linear transformation
T (f (x)) = xf ′(x) + f (1)
Is T diagonalizable? If it is, find a basis B such that [T ]B←B isdiagonal.Solution. Let D be the standard basis. Then
[T ]D←D︸ ︷︷ ︸A
=
1 1 1 ... 1
12
. . .n
.
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Characteristic Polynomial, Eigenvalue, Eigenvector and Diagonalization
Examples of Eigenvalues, Eigenvectors andDiagonalization
Solution (CONT). So the characteristic polynomial of T is(x − 1)2(x − 2)...(x − n).Note that
dim K (I − T ) = dim Nul(I − A) = 1dim K (2I − T ) = dim Nul(2I − A) = 1
... =...
dim K (nI − T ) = dim Nul(nI − A) = 1.
Therefore,
dim K (I−T )+dim K (2I−T )+...+dim K (nI−T ) = n < n+1 = dim Pn
and hence T is not diagonalizable.
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Characteristic Polynomial, Eigenvalue, Eigenvector and Diagonalization
Examples of Eigenvalues, Eigenvectors andDiagonalization
Let T : R2 → R2 be the linear transformation given by
T (x , y) = (3x + y ,4x + 3y).
Find T n and T 2014(1,1).Solution 1. We try to diagonalize T . Let C be the standardbasis. Then
[T ]C←C =
[3 14 3
]and det(xI − [T ]C←C) = x2 − 6x + 5.
So T has eigenvalues 1 and 5 with eigenvectors
K (I − T ) = Span{(1,−2)} and K (5I − T ) = Span{(1,2)}.Let B = {(1,−2), (1,2)}. Then
[T ]B←B =
[1
5
]Xi Chen Linear Algebra II Lecture 14
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Characteristic Polynomial, Eigenvalue, Eigenvector and Diagonalization
Examples of Eigenvalues, Eigenvectors andDiagonalization
Solution 1 (CONT). So
[T ]C←C = PC←B[T ]B←BP−1C←B
=
[1 1−2 2
] [1
5
] [1 1−2 2
]−1
Therefore,
[T n]C←C =
[1 1−2 2
] [1
5
]n [ 1 1−2 2
]−1
=14
[1 1−2 2
] [1
5n
] [2 −12 1
]=
14
[2(5n + 1) (5n − 1)4(5n − 1) 2(5n + 1)
]
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Characteristic Polynomial, Eigenvalue, Eigenvector and Diagonalization
Examples of Eigenvalues, Eigenvectors andDiagonalization
Solution 1 (CONT). So
T n(x , y) =14(2(5n + 1)x + (5n − 1)y ,4(5n − 1)x + 2(5n + 1)y)
andT 2014(1,1) =
14(3 · 52014 + 1,6 · 52014 − 2).
Solution 2. By Cayley-Hamilton, f (T ) = 0 for the characteristicpolynomial of T . Therefore, T 2 − 6T + 5I = 0. By long division,
T n = (T 2 − 6T + 5I)g(T ) + aT + bI ⇒ T n = aT + bI.
It suffices to find a and b.
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Characteristic Polynomial, Eigenvalue, Eigenvector and Diagonalization
Examples of Eigenvalues, Eigenvectors andDiagonalization
Solution 2 (CONT). Suppose that
xn ≡ (x2 − 6x + 5)g(x) + ax + b = (x − 1)(x − 5)g(x) + ax + b.
Let x = 1. Then a + b = 1.Let x = 5. Then 5a + b = 5n.Solve the system of linear equations:
{a + b = 1
5a + b = 5n ⇒
a =
14(5n − 1)
b =14(5− 5n)
T n = aT + bI ⇒ T n(x , y) = aT (x , y) + b(x , y)= ((3a + b)x + ay ,4ax + (3a + b)y).
Xi Chen Linear Algebra II Lecture 14