14.1.2 The Explicit Description Of The Method
Here is how you use this method to find the eigenvalue closest to α and the
- Pick u1. If you are not phenomenally unlucky, the iterations will converge.
- If uk has been obtained,
where sk+1 is the entry of
−1uk which has largest absolute value.
- When the scaling factors, sk are not changing much and the uk are not changing much, find
the approximation to the eigenvalue by solving
for λ. The eigenvector is approximated by uk+1.
- Check your work by multiplying by the original matrix to see how well what you have found
Thus this amounts to the power method for the matrix
but you are free to pick
Example 14.1.4 Find the eigenvalue of A =
which is closest to −
Also find an eigenvector which goes with this eigenvalue.
In this case the eigenvalues are −6,0, and 12 so the correct answer is −6 for the eigenvalue.
Then from the above procedure, I will start with an initial vector, u1 =
. Then I
must solve the following equation.
Simplifying the matrix on the left, I must solve
and then divide by the entry which has largest absolute value to obtain
and divide by the largest entry, 1.0515 to get
and divide by the largest entry, 1.01 to get
These scaling factors are pretty close after these few iterations. Therefore, the predicted eigenvalue
is obtained by solving the following for λ.
which gives λ = −6.01. You see this is pretty close. In this case the eigenvalue closest to −7 was
How would you know what to start with for an initial guess? You might apply Gerschgorin’s
theorem. However, sometimes you can begin with a better estimate.
Example 14.1.5 Consider the symmetric matrix A =
. Find the middle
eigenvalue and an eigenvector which goes with it.
Since A is symmetric, it follows it has three real eigenvalues which are solutions to
If you use your graphing calculator to graph this polynomial, you find there is an eigenvalue
somewhere between −.
9 and −.
8 and that this is the middle eigenvalue. Of course you could zoom
in and find it very accurately without much trouble but what about the eigenvector which goes
with it? If you try to solve
there will be only the zero solution because the matrix on the left will be invertible and the same
will be true if you replace −.8 with a better approximation like −.86 or −.855. This
is because all these are only approximations to the eigenvalue and so the matrix in
the above is nonsingular for all of these. Therefore, you will only get the zero solution
However, there exists such an eigenvector and you can find it using the shifted inverse power
method. Pick α = −.855. Then you solve
or in other words,
and after finding the solution, divide by the largest entry −67.944, to obtain
After a couple more iterations, you obtain
Then doing it again, the scaling factor is −513.42 and the next iterate is
Clearly the uk are not changing much. This suggests an approximate eigenvector for this
eigenvalue which is close to −.855 is the above u3 and an eigenvalue is obtained by
which yields λ = −0.85695 Lets check this.
Thus the vector of 14.4 is very close to the desired eigenvector, just as −.8569 is very close to the
desired eigenvalue. For practical purposes, I have found both the eigenvector and the
Example 14.1.6 Find the eigenvalues and eigenvectors of the matrix A =
This is only a 3×3 matrix and so it is not hard to estimate the eigenvalues. Just get the
characteristic equation, graph it using a calculator and zoom in to find the eigenvalues. If you do
this, you find there is an eigenvalue near −1.2, one near −.4, and one near 5.5. (The characteristic
equation is 2 + 8λ + 4λ2 − λ3 = 0.) Of course I have no idea what the eigenvectors
Lets first try to find the eigenvector and a better approximation for the eigenvalue near −1.2.
In this case, let α = −1.2. Then
As before, it helps to get things started if you raise to a power and then go from the approximate
Then the next iterate will be
Divide by largest entry
You can see the vector didn’t change much and so the next scaling factor will not be much
different than this one. Hence you need to solve for λ
Then λ = −1.2185 is an approximate eigenvalue and
is an approximate eigenvector. How well does it work?
You can see that for practical purposes, this has found the eigenvalue closest to −1.2185 and
the corresponding eigenvector.
The other eigenvectors and eigenvalues can be found similarly. In the case of −.4, you could let
α = −.4 and then
Following the procedure of the power method, you find that after about 5 iterations, the scaling
factor is 9.7573139, they are not changing much, and
Thus the approximate eigenvalue is
which shows λ = −.29751278 is an approximation to the eigenvalue near .4. How well does it
It works pretty well. For practical purposes, the eigenvalue and eigenvector have now been found.
If you want better accuracy, you could just continue iterating. One can find the eigenvector
corresponding to the eigenvalue nearest 5.5 the same way.