Find
λ
λ such that
Av=λv
A
v
λ
v
with
v≠0
v
0
.
Av−λv=0
A
v
λ
v
0
(1)
Av−λIv=0
A
v
λ
I
v
0
(2)
where
I
I is the indentity matrix in the form
1000010000......00...1
1
0
0
0
0
1
0
0
0
0
...
...
0
0
...
1
A−λIv=0
A
λ
I
v
0
(3)
This means, matrix
A−λI
A
λ
I
is
not invertible or
detA−λI=0
A
λ
I
0
(4)
where
det˙
˙
is the determinant.
So solve
detA−λI=0
A
λ
I
0
which will be a polynomial of
order
N
N;
λ
λ's are its roots.
A=3-1-13
A
3
-1
-1
3
A−λI=3-1-13−λ00λ=3−λ-1-13−λ
A
λ
I
3
-1
-1
3
λ
0
0
λ
3
λ
-1
-1
3
λ
(5)
detA−λI=det3−λ-1-13−λ=3−λ2−
-1
2=λ2−6λ+8=λ−2λ−4=0
A
λ
I
3
λ
-1
-1
3
λ
3
λ
2
-1
2
λ
2
6
λ
8
λ
2
λ
4
0
(6)
Therefore the eigenvalues are
λ
1
=2
λ
1
2
, and
λ
2
=4
λ
2
4
.
For
each
λ
i
λ
i
solve the
N×N
N
N
set of linear equations for
vi
v
i
Avi=λvi
A
v
i
λ
v
i
(7)
with
N
N equations in
N
N unknowns.
A=3-1-13
A
3
-1
-1
3
,
λ
1
=2
λ
1
2
,
λ
2
=4
λ
2
4
for
λ
1
=2
λ
1
2
,
3-1-13v11v12=2v11v12
3
-1
-1
3
v
11
v
12
2
v
11
v
12
(8)
v1=v11v12=11
v
1
v
11
v
12
1
1
(9)
for
λ
2
=4
λ
2
4
,
3-1-13v21v22=4v21v22
3
-1
-1
3
v
21
v
22
4
v
21
v
22
(10)
v2=v21v22=1-1
v
2
v
21
v
22
1
-1
(11)
What should be done with "repeated
eigenvalues" (ie: when
λ
i
=
λ
j
λ
i
λ
j
for some
λ
i
≠
λ
j
λ
i
λ
j
)?