MATH1231 Algebra, 2016
Chapter 8: Eigenvalues and eigenvectors
A/Prof. Daniel Chan
School of Mathematics and Statistics
University of New South Wales
[email protected]
A/Prof. Daniel Chan (UNSW)
MATH1231 Algebra
1 / 41
Motivating example: reflections
Example
Show that the linear map TA :
R2
−→
R2
associated to A =
0 1
1 0
is
reflection about some line.
Soln
To understand the map T , useful to change the co-ord axes to y = x and
y = −x. Then T is reflection about a co-ord axis.
Q Given A, what are good co-ordinates to understand TA ?
A The theory of eigenvalues and eigenvectors gives an answer.
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8.1 Introduction
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Definition of Eigenvectors and Eigenvalues
Definition (For Linear Maps)
Let V = vector space /F & T : V → V be linear. Then if a scalar λ ∈ F
and a non-zero vector v ∈ V satisfy
T (v) = λv,
then λ ∈ F is called an eigenvalue of T and v is called an eigenvector of
T for the eigenvalue λ.
Note: The domain and codomain are the same vector space.
Definition (For Matrices)
Let A be an n × n square matrix. Then if a scalar λ ∈ F and non-zero
vector x ∈ Fn satisfy
Ax = λx,
then λ is called an eigenvalue of A and x is called an eigenvector of A
for the eigenvalue λ.
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8.1 Introduction
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Eigenvalues for reflection
Example
0 1
Some e-vectors & e-values for the reflection matrix A =
.
1 0
1
1
1
1
=1
∴
is an e-vector with e-value 1.
A
=
1
1
1
1
1
1
−1
=−
∴ is an e-vector with e-value − 1.
A
=
−1
−1
1
Remark
1
Any non-zero vector on line of reflection y = x is an e-vector with
e-value 1.
2
Sim, any non-zero vector on the orthogonal line y = −x is an e-vector
with e-value -1.
3
No other e-vectors so 1, −1 are the only e-values for A.
4
Upshot Eigenvectors give the good co-ord axes!
A/Prof. Daniel Chan (UNSW)
8.1 Introduction
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Eigenspaces
Q How do you find e-vectors & e-values?
Partial A Given λ ∈ F, find e-vectors with e-value λ using:
Theorem-Defn
The λ-eigenspace of a square matrix A is the subspace ker(A − λI ). The
e-vectors of A with e-value λ are precisely the non-zero vectors of
ker(A − λI ).
Proof.
v 6= 0 is an e-vector with e-value λ ⇐⇒
Corollary
λ is an e-value for A iff ker(A − λI ) 6= 0.
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8.1 Introduction
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Examples & Remarks on e-vectors
0 1
E.g. The 1-eigenspace of the reflection matrix A =
is
1 0
ker(A − 1I ) =
Rem
The 0-e-space is just kerA so the e-vectors with e-value 0 are the
non-zero vectors in kerA.
ker(A − λI ) is closed under scalar multn i.e. v an e-vector with
e-value λ =⇒ so is any non-zero scalar multiple of v.
A/Prof. Daniel Chan (UNSW)
8.1 Introduction
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Finding e-values
To find e-values, we use
Theorem
λ is an e-value of a square matrix A iff det(A − λI ) = 0.
Proof.
λ is an e-value for A ⇐⇒ ker(A − λI ) 6= 0 ⇐⇒
solns to (A − λI )x = 0 are not unique ⇐⇒ (A − λI ) is not invertible ⇐⇒
det(A − λI ) = 0.
A/Prof. Daniel Chan (UNSW)
8.1 Introduction
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2 × 2 Matrices
Example
Find the eigenvalues and eigenvectors of the matrix
3 −2
.
−1
2
Solution
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8.1 Introduction
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Solution (Continued)
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8.1 Introduction
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Solution (Continued)
This
example,
e-values are distinct. Have a basis of e-vectors
2
1
,
which gives good co-ords to study A.
−1
1
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8.1 Introduction
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Characteristic polynomial
Theorem
Let A = n × n matrix /C. Then det(A − λI ) is a polynomial of degree
n in λ. The polynomial det(A − λI ) is called the characteristic
polynomial for A.
An n × n matrix always has n e-values in C (not necessarily distinct).
If all n e-values are distinct, and B = {v1 , . . . , vn } are e-vectors with
distinct e-values, then B is a basis for Cn (and for Rn if the vectors
are real).
Warning If e-values are not distinct
then
there may not be a basis of
1 1
e-vectors. This happens for A =
.
0 1
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8.1 Introduction
11 / 41
Example
2 0
Find the eigenvalues and eigenvectors of the matrix
.
0 2
Solution
Hence, the eigenvalue is 2.
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8.1 Introduction
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Solution (Continued)
Compute 2-e-space ker(A − 2I ). Solve
0 0
0
(A − 2I )v = 0 ⇐⇒
v=
.
0 0
0
v can be anything i.e. 2-e-space is R2 .
Hence any non-zero vector in R2 is an e-vector. The e-vectors are
1
0
s
+t
,
0
1
not both s and t are 0.
1
0
Here the e-values are equal. Still have a basis of e-vectors
,
.
0
1
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8.1 Introduction
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Example
Find the eigenvalues and eigenvectors of the matrix
1 1
.
−1 1
Solution
1 1
Let A =
.
−1 1
E-values of A are the solns to characteristic eqn
det(A − λI )
1−λ
1
det
−1 1 − λ
(1 − λ)(1 − λ) − (−1)
=0
2
(λ − 1)
=0
=0
= −1
λ=1±i
E-values are 1 + i and 1 − i.
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8.1 Introduction
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Solution (Continued)
v
The e-vectors v = 1 of A for λ = 1 + i are non-zero solns to
v2
(A − (1 + i)I )v = 0
1 − (1 + i)
1
−i
1
0
which is
v=
v=
.
−1
1 − (1 + i)
−1 −i
0
Reduce the augmented matrix with the right hand zero column omitted:
−i
1
−1 −i
R2 = R2 + iR1
−i 1
−−−−−−−−−−−−−→
0 0
Hence
−iv1 + v2 = 0. Put v2 = t so v1 = −it. E-vectors are
−i
t
, t 6= 0.
1
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8.1 Introduction
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Solution (Continued)
v
E-vectors v = 1 of A for λ = 1 − i are non-zero solns to
v2
(A − (1 − i)I )v = 0
1 − (1 − i)
1
i 1
0
which is
v=
v=
.
−1
1 − (1 − i)
−1 i
0
Reduce the augmented matrix with the right hand zero column omitted:
R2 = R2 − iR1
i 1
i 1
−−−−
−−−−−−−−−→
−1 i
0 0
i
Hence iv1 + v2 = 0. E-vectors are t
, t 6= 0.
1
Here,
e-values
are distinct & not real. Again have basis of e-vectors
−i
i
,
.
1
1
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8.1 Introduction
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Higher Order Square Matrices
Example
Find the eigenvalues and eigenvectors of the matrix


−3 4
2 −3
 2 12 −4
2
.
A=
 −5 12
2 −1 
−15 4
2
9
Solution (Use Maple)
with(LinearAlgebra):
A := <<-3, 2, -5, -15>|<4, 12, 12, 4>|<2, -4, 2, 2>|<-3,
2, -1, 9>>;


−3 4
2 −3
 2
12 −4 2 


 −5 12 2 −1 
−15 4
2
9
>
>
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8.1 Introduction
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Solution (Continued)
Eigenvectors(A);
 


1 1 0 1
4
 


 −4   1 0 1 2 
2
 


, 


1
 12   3 1

2 3 
 

8
1 1 1 1
Check these directly!
>
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8.1 Introduction
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Diagonal matrices are easy to multiply
Multiplication of diagonal matrices is easy.




c1 0 . . . 0
d1 0 . . . 0
 0 c2
 0 d2
0
0




C =.
,
D
=


.
.
.. 
.
.
.
.
.
.
.
.
.
. .
. .
0 0 . . . cn
0 0 . . . dn
Proposition


 k
d1
c 1 d1
0
...
0
 0


c2 d2
0 

0
i) CD =  .
.. , ii) D k =  ..
.
.
.
 .
 .
.
. 
0
0
. . . cn dn
0
0
d2k
0
...
..

0
0

.. 
. 
.
. . . dnk
Note that i) =⇒ ii) by induction.
A/Prof. Daniel Chan (UNSW)
8.2 Diagonalisation
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Examples of multiplying diagonal matrices
Propn clear from any eg
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8.2 Diagonalisation
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Diagonalisation: motivation
Key Fact A square matrix A can be recovered from a basis of e-vectors &
corresponding e-values.
Example
Let A = 3 × 3 matrix with e-values 1, 3, −2 & the corresponding basis of
e-vectors v1 , v2 , v3 . Find A.


1 0
0
0 .
Let M = (v1 |v2 |v3 ) and D = 0 3
0 0 −2
AM = A(v1 |v2 |v3 ) = (Av1 |Av2 |Av3 ) = (v1 |3v2 | − 2v3 )


1 0
0
0  = MD.
= (v1 |v2 |v3 ) 0 3
0 0 −2
Columns of M are lin indep =⇒ M is invertible.
∴ M −1 AM = D or A = MDM −1 .
A/Prof. Daniel Chan (UNSW)
8.2 Diagonalisation
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Diagonalisation
Theorem (Diagonalisation)
Suppose n × n-matrix A has a basis {v1 , . . . , vn } of e-vectors with
corresponding e-values λ1 , . . . , λn . Let M = (v1 |v2 | . . . |vn ) &
D = (λ1 e1 | . . . |λn en ) be the diagonal matrix with i-th diagonal entry λi .
Then
M −1 AM = D.
Conversely if M −1 AM = D with D diagonal then
the columns of M are a basis of e-vectors of A &
the diagonal entries of D give the corresponding e-values.
Definition
A square matrix A is diagonalisable if there exists an invertible matrix M
and diagonal matrix D such that M −1 AM = D.
A/Prof. Daniel Chan (UNSW)
8.2 Diagonalisation
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Diagonalisation: example
Example
3 −2
Diagonalise A =
i.e. Find an invertible matrix M & diagonal
−1
2
matrix D, so that M −1 AM = D.
2
1
Soln We’ve already seen
,
are e-vectors with e-values
−1
1
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8.2 Diagonalisation
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Diagonalisation: another example
Example
1 1
Show A =
is not diagonalisable by showing it does not have a
0 1
basis of e-vectors.
Solution
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8.2 Diagonalisation
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Example cont’d
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8.2 Diagonalisation
25 / 41
Powers of a Matrix
Suppose
M −1 AM = D or A = MDM −1 .
Then
An = MDM −1
n
= MDM −1
MDM −1 · · · MDM −1 = MD n M −1 .
Easy to compute if D is diagonal!
Example
3 −2
Let A =
. Find A100 and An .
−1
2
Solution
−1
We use
M
the diagonalisation
AM = D where
2 1
4 0
M=
,D =
.
−1 1
0 1
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8.3 Applications
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Example cont’d: powers of a matrix
Solution (Continued)
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8.3 Applications
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Example cont’d: powers of a matrix
Solution (Continued)
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8.3 Applications
28 / 41
Why diagonal matrices are good: decoupled equations
d1 0
Consider diagonal D =
. Eqn Dx = b is really easy to solve ∵
0 d2
it’s
d1 x1
= b1
d2 x2 = b2
The 1st eqn only involves x1 whilst the 2nd only involves x2 .
∴ can solve the eqns separately & we call these eqns decoupled.
Upshot
row-echelon form =⇒ easy to solve, but
diagonal form =⇒ REALLY, REALLY easy to solve.
Remark The same is true of differential equations! as we will see on the
next slide.
A/Prof. Daniel Chan (UNSW)
8.3 Applications
29 / 41
Decoupled ODEs
Suppose that a population of hobbits at time t is x(t) & the population of
orcs is y (t). If they live separately, say in the Shire & in Mordor, the
populations grow independently according to a DE like
Example

dx


= 3x

dt


 dy =
dt
.
2y
Soln These are decoupled & we can solve for x, y separately
x(t) = αe 3t ,
A/Prof. Daniel Chan (UNSW)
y (t) = βe 2t .
8.3 Applications
30 / 41
Matrix form for (de)coupled ODEs
Let’s now put the hobbits and orcs together in New Zealand. Then they
evolve according to a coupled ODE like
Example

dx


= 3x − 2y

dt
.

dy


= − x + 2y
dt
3 −2
x
and A =
, so
We rewrite in matrix form. Let y =
y
−1
2
 
dx

3x − 2y
3
−2
dt 
0


y =  =
=
y =⇒ y0 = Ay.
−x
+
2y
−1
2
dy
dt
The fact that the ODE is coupled corresponds to the fact that A is not
diagonal.
A/Prof. Daniel Chan (UNSW)
8.3 Applications
31 / 41
Solving ODEs by decoupling
Consider diagonalised matrix A = MDM −1 & ODE
Change variables to z = M −1 y & use
dy
dt
= Ay.
Proposition
Given a constant matrix C we have
d
dt
(C y) = C dy
dt .
Since M −1 A = DM −1 we have
dz
d
dy
=
M −1 y = M −1
= M −1 Ay = DM −1 y = Dz
dt
dt
dt
If λ1 , . . . , λn are the diagonal entries of D, then this decoupled ODE in z
can be solved as in the hobbits in the Shire/ orcs in Mordor example




α1 e λ1 t
α 1 e λ1 t




z =  ...  =⇒ y = M  ... 
α n e λn t
αn e λn t
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8.3 Applications
32 / 41
General solution to
dy
dt
= Ay: explicit formula
If M = (v1 | . . . |vn ), we can multipy matrices to get
Theorem
Let A be an n × n matrix with a basis of e-vectors v1 , . . . , vn &
corresponding e-values λ1 , . . . , λn . Then the general solution to y0 = Ay is
y(t) = α1 e λ1 t v1 + · · · + αn e λn t vn
for arbitrary constants α1 , . . . , αn ∈ R.
Note It’s easy to check the expression above does give a solution.
A/Prof. Daniel Chan (UNSW)
8.3 Applications
33 / 41
Back to hobbits and orcs
Example
Recall the hobbit/ orc population in NZ is governed by
3 −2
A=
.
−1
2
1 Find the general soln to the ODE.
2
dy
dt
= Ay where
Find the population as a function of time if the initial population
consisted of 4000 hobbits and 1000 orcs.
Solution
We found e-vectors for A:
A/Prof. Daniel Chan (UNSW)
2
−1
1
,
with coresponding e-values 4, 1.
1
8.3 Applications
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Hobbit/orc example cont’d
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8.3 Applications
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Alternate method via change of variable & decoupling
2 1
4 0
,D =
.
−1 1
0 1
Change
to z = M −1 y. Slide 32 gives decoupled ODE
variables
4 0
dz
dt = 0 1 z i.e.
dz1
dz2
= 4z1 ,
= z2 .
dt
dt
z1 (t)
α1 e 4t
General soln z =
=
, α1 , α2 ∈ R.
α2 e t
z2 (t)
To find integration constants α1 , α2 , we use the initial value for z i.e.
z(0) = M −1 y(0) which can be found by solving
4000
Mz(0) = y(0) =
.
1000
Recall D = M −1 AM where M =
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8.3 Applications
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Alternate method cont’d
2 1 4000
−1 1 1000
R2=R2+ 12 R1
−−−−−−−−→
2 1 4000
0 32 3000
so z2 (0) = 32 × 3000 = 2000 &
2z1 (0) + 1 × 2000 = 4000 =⇒ z1 (0) = 1000.
Thus
α1
1000e 4t
z1 (0)
1000
= z(0) =
=
=⇒ z =
.
α2
2000e t
z2 (0)
2000
We go back to original variables
2 1
1000e 4t
2000e 4t + 2000e t
=
.
y = Mz =
−1 1
2000e t
−1000e 4t + 2000e t
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8.3 Applications
37 / 41
Solving 2nd order ODEs using systems of 1st order ODEs
Example
Solve y 00 + 4y 0 − 5y = 0 by converting this ODE into a system of first
order differential equations.
Solution
The trick is to let
y1 = y
and
y2 = y10 = y 0 ,
y
and y = 1 . We have y 00 = 5y − 4y 0 = 5y1 − 4y2 so
y2
0 y1
y2
0
1
y =
=
=
y.
y20
5y1 − 4y2
5 −4
0
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8.3 Applications
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Solution (Continued)
0
1
E-values of A =
are the solns to the characteristic eqn
5 −4
det(A − λI ) = 0
−λ
1
det
=0
5 −4 − λ
−λ(−4 − λ) − 5 = 0
λ2 + 4λ − 5 = 0
(λ − 1)(λ + 5) = 0
λ = 1, −5.
A/Prof. Daniel Chan (UNSW)
8.3 Applications
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Solution (Continued)
v1
of A for e-value 1 are non-zero solns to
v2
−1
1
v1
0
(A − I )v = 0 i.e.
=
5 −5
v2
0
1
Hence −v1 + v2 = 0. Pick e-vector v =
.
1
Sim, solve (A + 5I )v = 0 to get an e-vector for λ = −5.
5 1
v1
0
=
i.e. 5v1 + v2 = 0.
5 1
v2
0
1
Pick e-vector v =
for λ = −5. General soln is
−5
E-vectors v =
y=α
1
1
et + β
e −5t .
−5
1
Hence y = y1 = αe t + βe −5t for constants α, β.
A/Prof. Daniel Chan (UNSW)
8.3 Applications
40 / 41
Initial Value Problem
Example
Solve the IVP y 00 + 4y 0 − 5y = 0, y (0) = 1, y 0 (0) = −5.
Solution
Use gen soln y = α
1
−5
1
1
et + β
e −5t . To find α, β note
1
−5
y (0)
=
y 0 (0)
1
1
+β
= y(0) = α
1
−5
You can now solve by Gaussian elimination, or better still note
α = 0, β = 1 must be the soln! Hence
1
y=
e −5t =⇒ y = y1 = e −5t .
−5
A/Prof. Daniel Chan (UNSW)
8.3 Applications
41 / 41