Assume that the n×n matrix A has distinct w:eigenvalues
λ
1
{\displaystyle \lambda _{1}}
,
λ
2
{\displaystyle \lambda _{2}}
,....
λ
n
{\displaystyle \lambda _{n}}
and consider the eigenvalue
λ
j
{\displaystyle \lambda _{j}}
.
Then a constant α can be chosen so that
σ
1
{\displaystyle \sigma _{1}}
=1 / (
λ
j
{\displaystyle \lambda _{j}}
- α)
is the dominant eigenvalue of
(
A
−
α
I
)
−
1
{\displaystyle (A-\alpha I)^{-1}}
.
Furthermore, if
X
0
{\displaystyle X_{0}}
, which is the initial guess vector, is chosen appropriately , then the sequences
(
X
k
)
{\displaystyle \left(X_{k}\right)}
defined by
Y
k
=
(
A
−
α
I
)
−
1
X
k
{\displaystyle Y_{k}=(A-\alpha I)^{-1}X_{k}}
X
k
+
1
=
Y
k
‖
Y
k
‖
{\displaystyle X_{k+1}={\frac {Y_{k}}{\|Y_{k}\|}}}
and
(
c
k
+
1
)
{\displaystyle \left(c_{k+1}\right)}
defined by
c
k
+
1
=
Y
k
⊤
X
k
X
k
⊤
X
k
{\displaystyle c_{k+1}={\frac {Y_{k}^{\top }X_{k}}{X_{k}^{\top }X_{k}}}}
(w:Rayleigh quotient )
will converge to the dominant eigenpair
σ
1
{\displaystyle \sigma _{1}}
,
X
k
+
1
{\displaystyle X_{k+1}}
will converge to the corresponding eigenvector
V
1
{\displaystyle V_{1}}
of the matrix
(
A
−
α
I
)
−
1
{\displaystyle (A-\alpha I)^{-1}}
.
Therefore, the corresponding eigenvalue for the matrix A is given by
λ
j
=
1
/
σ
1
+
α
{\displaystyle \lambda _{j}=1/\sigma _{1}+\alpha }
.
Use the shifted inverse power method to find the eigenpairs of the matrix
A
=
[
0
11
−
5
−
2
17
−
7
−
4
26
−
10
]
{\displaystyle A=\left[{\begin{array}{c c c}0&11&-5\\-2&17&-7\\-4&26&-10\end{array}}\right]}
.
Use the fact that the eigenvalues of A are
λ
1
{\displaystyle \lambda _{1}}
=4,
λ
2
{\displaystyle \lambda _{2}}
=2,
λ
3
{\displaystyle \lambda _{3}}
=1,
and select an appropriate α and starting vector for each case.
Case1: For the eigenvalue
λ
1
{\displaystyle \lambda _{1}}
=4,
we select α=4.2 and the starting vector
X
0
=
[
1
1
1
]
{\displaystyle X_{0}=\left[{\begin{array}{c}1\\1\\1\\\end{array}}\right]}
.
First we can get
(
A
−
4.2
I
)
=
[
−
4.2
11
−
5
−
2
12.8
−
7
−
4
26
−
14.2
]
{\displaystyle (A-4.2I)=\left[{\begin{array}{c c c}-4.2&11&-5\\-2&12.8&-7\\-4&26&-14.2\end{array}}\right]}
and then we can apply the shifted inverse power method
Y
k
{\displaystyle Y_{k}}
=
(
A
−
α
I
)
−
1
{\displaystyle (A-\alpha I)^{-1}}
X
k
{\displaystyle X_{k}}
.
Therefore,
[
−
4.2
11
−
5
−
2
12.8
−
7
−
4
26
−
14.2
]
{\displaystyle \left[{\begin{array}{c c c}-4.2&11&-5\\-2&12.8&-7\\-4&26&-14.2\end{array}}\right]}
Y
0
{\displaystyle Y_{0}}
=
X
0
=
[
1
1
1
]
{\displaystyle X_{0}=\left[{\begin{array}{c}1\\1\\1\\\end{array}}\right]}
.
Solving this system of equations, we get
Y
0
=
[
−
9.545454545
−
14.09090909
−
23.18181818
]
{\displaystyle Y_{0}=\left[{\begin{array}{c}-9.545454545\\-14.09090909\\-23.18181818\\\end{array}}\right]}
.
Next we can compute
c
1
{\displaystyle c_{1}}
=
Y
0
⊤
X
0
X
0
⊤
X
0
.
{\displaystyle {\frac {Y_{0}^{\top }X_{0}}{X_{0}^{\top }X_{0}}}.}
,
so
c
1
{\displaystyle c_{1}}
=-15.606060605.
Since
X
k
+
1
=
Y
k
‖
Y
k
‖
.
{\displaystyle X_{k+1}={\frac {Y_{k}}{\|Y_{k}\|}}.}
,
it implies
X
1
=
[
−
0.4117
−
0.6078
−
1
]
{\displaystyle X_{1}=\left[{\begin{array}{c}-0.4117\\-0.6078\\-1\\\end{array}}\right]}
.
We continue doing the second iteration:
[
−
4.2
11
−
5
−
2
12.8
−
7
−
4
26
−
14.2
]
{\displaystyle \left[{\begin{array}{c c c}-4.2&11&-5\\-2&12.8&-7\\-4&26&-14.2\end{array}}\right]}
Y
1
{\displaystyle Y_{1}}
=
X
1
=
[
−
0.4117
−
0.6078
−
1
]
{\displaystyle X_{1}=\left[{\begin{array}{c}-0.4117\\-0.6078\\-1\\\end{array}}\right]}
.
Thus
Y
1
=
[
2.14795
3.21746
5.35650
]
{\displaystyle Y_{1}=\left[{\begin{array}{c}2.14795\\3.21746\\5.35650\\\end{array}}\right]}
.
It implies
c
2
{\displaystyle c_{2}}
=-5.326069
and
X
2
=
[
0.400998
0.600665
1
]
{\displaystyle X_{2}=\left[{\begin{array}{c}0.400998\\0.600665\\1\\\end{array}}\right]}
.
We should continue the iteration and finally we got the sequence
(
c
k
)
{\displaystyle \left(c_{k}\right)}
will converge to
σ
1
=
−
5
{\displaystyle \sigma _{1}=-5}
,
which is the dominant eigenvalue of
(
A
−
4.2
I
)
−
1
{\displaystyle (A-4.2I)^{-1}}
,
and the sequences
(
X
k
)
{\displaystyle \left(X_{k}\right)}
converges to
V
1
=
[
0.4
0.6
1
]
{\displaystyle V_{1}=\left[{\begin{array}{c}0.4\\0.6\\1\\\end{array}}\right]}
after 9 iterations. We can get the eigenvalue
λ
1
{\displaystyle \lambda _{1}}
of A by the formula:
λ
1
{\displaystyle \lambda _{1}}
= 1 /
σ
1
{\displaystyle \sigma _{1}}
+ α= 1/(-5) + 4.2 =4.
We can apply the same approach to find another two eigenvalues of the given matrix A.
Use the shifted inverse power method to find the eigenvalue
λ
2
{\displaystyle \lambda _{2}}
=2
for the same matrix A as the example above, given the starting vector
X
0
=
[
1
1
1
]
{\displaystyle X_{0}=\left[{\begin{array}{c}1\\1\\1\\\end{array}}\right]}
,
α=2.1.
For the eigenvalue
λ
1
{\displaystyle \lambda _{1}}
=2,
we select α=2.1 and the starting vector
X
0
=
[
1
1
1
]
{\displaystyle X_{0}=\left[{\begin{array}{c}1\\1\\1\\\end{array}}\right]}
.
First we can get
(
A
−
2.1
I
)
=
[
−
2.1
11
−
5
−
2
14.9
−
7
−
4
26
−
12.1
]
{\displaystyle (A-2.1I)=\left[{\begin{array}{c c c}-2.1&11&-5\\-2&14.9&-7\\-4&26&-12.1\end{array}}\right]}
.
Therefore,
[
−
2.1
11
−
5
−
2
14.9
−
7
−
4
26
−
12.1
]
{\displaystyle \left[{\begin{array}{c c c}-2.1&11&-5\\-2&14.9&-7\\-4&26&-12.1\end{array}}\right]}
Y
0
{\displaystyle Y_{0}}
=
X
0
=
[
1
1
1
]
{\displaystyle X_{0}=\left[{\begin{array}{c}1\\1\\1\\\end{array}}\right]}
.
So
Y
0
=
[
11.05263158
21.57894737
42.63157895
]
{\displaystyle Y_{0}=\left[{\begin{array}{c}11.05263158\\21.57894737\\42.63157895\\\end{array}}\right]}
and
c
1
=
25.0877193
{\displaystyle c_{1}=25.0877193}
It implies
X
1
=
[
0.2592592593
0.5061728395
1
]
{\displaystyle X_{1}=\left[{\begin{array}{c}0.2592592593\\0.5061728395\\1\\\end{array}}\right]}
.
After 7 iterations,we got
σ
1
=
−
10
{\displaystyle \sigma _{1}=-10}
and
V
1
=
[
0.25
0.5
1
]
{\displaystyle V_{1}=\left[{\begin{array}{c}0.25\\0.5\\1\\\end{array}}\right]}
.
Doing some computation, We got
λ
1
{\displaystyle \lambda _{1}}
= 1 /
σ
1
{\displaystyle \sigma _{1}}
+ α= 1/(-10) + 2.1 =2.
Use w:Matlab to do the shifted inverse power method to find the eigenvalue
λ
2
{\displaystyle \lambda _{2}}
=5.1433
for the given matrix
A
=
[
6
2
−
1
2
5
1
−
1
1
4
]
{\displaystyle A=\left[{\begin{array}{c c c}6&2&-1\\2&5&1\\-1&1&4\end{array}}\right]}
.
The starting vector is
X
0
=
[
1
1
3
]
{\displaystyle X_{0}=\left[{\begin{array}{c}1\\1\\3\\\end{array}}\right]}
,
α=6.
First we can get
(
A
−
6
I
)
=
[
0
2
−
1
2
−
1
1
−
1
1
−
2
]
{\displaystyle (A-6I)=\left[{\begin{array}{c c c}0&2&-1\\2&-1&1\\-1&1&-2\end{array}}\right]}
.
Then we can apply the method mentioned above to find the middle eigenvalue of the matrix. Below is the Matlab code for this question.
for i = 1 : 7
y = linsolve ( A , x );
e =( y '* x ) / ( x '* x );
x = y / norm ( y );
end
We can find that the sequence
(
c
k
)
{\displaystyle \left(c_{k}\right)}
will converge to
σ
2
=
−
1.167238857441354
{\displaystyle \sigma _{2}=-1.167238857441354}
after 7 iterations.
Doing some computation, We got
λ
2
{\displaystyle \lambda _{2}}
= 1 /
σ
2
{\displaystyle \sigma _{2}}
+ α= 1/(-1.167238857441354) + 6 = 5.143277321839636
which is approximately equal to 5.1433, the middle eigenvalue of the matrix.