Eigenvalues

1. The world of Eigenvalues-eigenfunctions
An operator A operates on a function and produces a
function.
For every operator, there is a set of functions which
when operated by the operator produces the same
function modified only multiplied by a constant
factor.
Such a function is called the eigenfunction of the
operator, and the constant modifier is called its
corresponding eigenvalue. An eigenvalue is just a
number: Real or complex.
A typical eigenvalue equation would look like
xAxl=
Here, the matrix or the operator A operates on a
vector (or a function) x producing an amplified or
reduced vector xl. Here the eigenvalue lbelongs to
eigenfunction x.
Suppose the operator is
)(
dx
d
xA=. A operating on
n
x produces
nnn
nxx
dx
d
xAx == .

2. Therefore, the operator A has an eigenvalue n
corresponding to eigenfunction
n
x.
1.Eigenfunctions are not unique.
Suppose xAxl=. Define, another vector cxz=, where
c is a constant.
Now, zcxxccAxAcxAz lll =====
Therefore, z is also an e-function (eigenfunction)
of .A
2.If xAxl= is an eigenvalue equation (and we
assume that x is not a zero vector), then
0I)-det(A 0I)x-(A xAx =ÜÞ=Û= lll
This leads to a characteristic polynomial in l:
)IAdet(p
A l-=
l is an e-value of A only if
0=
A
p.
3.Spectrum of an operator A is
)A(s: set of all its
e-values.
4.Spectral radius of an operator A is

||max)A(
)A(
lr
slÎ
=
=
||max
i
ni
l
££1
5.Computation of spectrum and spectral radius:

3. Let ú
û
ù
ê
ë
é-
=
52
12
A be the matrix and we want to
compute its eigenvalues and eigenfunctions. Its
characteristic equation (CE) is:
02)-)(5-(2 det =+ÜÞ=
ú
û
ù
ê
ë
é
-
--
ll
l
l
0
52
12
This gives 0127
2
=+-ll Ü Þ
043 =-- ))(( ll
Therefore, A has two eigenvalues: 3 and 4.
Let the eigenfunction be the vector ú
û
ù
ê
ë
é
=
2
1
x
x
x
corresponding to e-value 3. Then
ú
û
ù
ê
ë
é
=
ú
û
ù
ê
ë
é
=
ú
û
ù
ê
ë
é
ú
û
ù
ê
ë
é-
2
1
2
1
2
1
3
3
3
52
12
x
x
x
x
x
x
Therefore, we have 121
32 xxx =- yielding
21
xx-=. Also, we get 221
352 xxx =+ which gives us no new
result. Therefore, we can arbitrarily take the
following solution: ú
û
ù
ê
ë
é
-
=
1
1
1
e corresponding to e-value 3
for the matrix A.

4. Similarly, for e-value of 4, the eigenfunction appears
to be ú
û
ù
ê
ë
é
-
=
2
1
2e .
6.Faddeev-Leverrier Method to get characteristic
polynomial.
Define a sequence of matrices
)P(tracep ,AP
1 11 ==
[ ] )P(tracep ,IpPAP
2 2112
2
1
=-=
[ ] )P(tracep ,IpPAP
3 3223
3
1
=-=
…
…
[ ] )P(trace
n
p ,IpPAP
nnnnn
1
11 =-=
--
Then the characteristic polynomial
)(Pl is
[ ]
n
nnnn
p...pp)()(P -----=
-- 2
2
1
1
1 llll
e.g.
ú
ú
ú
û
ù
ê
ê
ê
ë
é
-
-
=
1626
2166
6612
A
Define
44161612
1 =++=== )A(tracep ,AP
1
=-= )IpP(AP
112
ú
ú
ú
û
ù
ê
ê
ê
ë
é
--
-
--
ú
ú
ú
û
ù
ê
ê
ê
ë
é
-
-
2826
2286
6632
1626
2166
6612
564
40860108
60408108
108108312
-=
ú
ú
ú
û
ù
ê
ê
ê
ë
é
--
---
--
=
2
p ,

5. And one proceeds this way to get
1728
3
=p
The CA polynomial = [ ]1728564441
233
-+-- lll)(
The eigenvalues are next found solving
[ ]0172856444
23
=-+- lll
7.More facts about eigenvalues.
Assume xAxl=. Therefore, l is the eigenvalue of
A with eigenvector x.
a.
1-
A has the same eigenvector as A and the
corresponding eigenvalue is
1-
l.
b.
n
A has the same eigenvector as A with the
eigenvalue
n
l.
c.
)IA(m+ has the same eigenvector as A with the
eigenvalue
)(ml+.
d. If A is symmetric, all its eigenvalues are real.
e. If P is an invertible matrix then APP
1-
has the
same eigenvalues as A.
Proof of e.

6. Suppose, the eigenfunction of APP
1-
is
y
with
eigenvalue k.
Then,
kyAPyP =
-1
Ü Þ
kPyPkyAPy ==
Therefore,
xPy= and k must be equal to l. Therefore
the eigenvalues of A and APP
1-
are identical and the
eigenvector of one is a linear mapping of the other
one.
If the eigenvalues of A, n
,...,,lll
21 are all distinct
then there exists a similarity transformation such that
ú
ú
ú
ú
ú
ú
û
ù
ê
ê
ê
ê
ê
ê
ë
é
==
-
n
..
........
..
..
..
DAPP
l
l
l
l
000
0
000
000
000
3
2
1
1

Let the eigenvectors of A be
)n()i()()(
x,...x,...,x,x
21
such that we have
)i(
i
)i(
xAxl=
Then the matrix [ ]
)n()()(
x,...,x,xP
21
=
Then [ ]
)n()()(
Ax,...,Ax,AxAP
21
=
[ ]
)n(
n
)()(
x,...,x,x lll
2
2
1
1
=
[ ][ ]
)n(
n
)()()n()()(
e,...,e,ex,...,x,x lll
2
2
1
1
21
=
PD=
Therefore, DAPP =
-1
Also, note the following. If A is symmetric, then

7. ji ,x)x(
)j(t)i(
¹"=0. So, we can normalize each
eigenvector and obtain )i(
)i(
)i(
x
x
u=
so that the
matrix [ ]
)n()()(
u,...,u,uQ
21
= would be an orthogonal matrix.
i.e.
DAQQ
t
=
Matrix-norm .
Computationally, the 2
l-norm of a matrix is
determined as
2
l-norm of [ ]
21
2
/
t
)AA(||A||A r==
e.g.
ú
ú
ú
û
ù
ê
ê
ê
ë
é
-
=
211
121
011
A
Then
ú
ú
ú
û
ù
ê
ê
ê
ë
é
-
-
=
ú
ú
ú
û
ù
ê
ê
ê
ë
é
-ú
ú
ú
û
ù
ê
ê
ê
ë
é -
=
541
462
123
211
121
011
210
121
111
AA
t
The eigenvalues are:
77770
1 -=+==
32 , , lll
Therefore,
106377
2
.)AA(A
t
»+==r
The ¥lnorm is defined as
å
££
¥
=
j
ij
ni
amaxA
1
e.g.
ú
ú
ú
û
ù
ê
ê
ê
ë
é
--
=
411
121
011
A

8. 2011
3
1
1 =++=å
=j
ja
,
4121
3
1
2 =++=å
=j
ja
6
3
1
3
=å
=j
j
a
Therefore,
6642==
¥
),,max(A
In computational matrix algebra, we would often be
interested about situations when
k
A becomes small
(all the entrees become almost zero). In that case, Ais
considered convergent.
i.e. A is convergent if
( )0=
¥®
ij
k
k
Alim

Example. Is
ú
ú
ú
û
ù
ê
ê
ê
ë
é
=
2
1
4
1
0
2
1
A convergent?
ú
ú
ú
û
ù
ê
ê
ê
ë
é
=
4
1
4
1
0
4
1
2
A ,
ú
ú
ú
û
ù
ê
ê
ê
ë
é
=
8
1
16
3
0
8
1
3
A ,
ú
ú
ú
û
ù
ê
ê
ê
ë
é
=
16
1
8
1
0
16
1
4
A ,
It appears that
ú
ú
ú
ú
û
ù
ê
ê
ê
ê
ë
é
+
=
kk
k
k
k
A
2
1
12
0
2
1

9. In the limit ¥®k,
0
2
1
®
k. Therefore, A is a convergent
matrix.
Note the following equivalent results:
a. A is a convergent matrix
b1.
0
2
=
¥®
k
k
Alim

b2.
0=
¥¥®
k
k
Alim
c.
1<)A(r
d.
x xAlim
k
k
"=
¥®
0
Condition number
)A(K of a non-singular matrix A
is computed as
-1
A . A)A(K=
A matrix is well-behaved if its condition number is
close to 1. When
)A(K of a matrix A is significantly
larger than 1, we call it an ill-behaved matrix.