If you have a good method that can easily find the largest eigenvalue of A, then you can use the same method to find smallest eigenvalue of A by finding the largest eigenvalue of -A and multiplying by -1.

Are there methods to find the smallest or largest eigenvalue tout court? I thought that these methods can only find the eigenvalue of smallest or largest modulus.

I would recommend looking up the wikipedia chapter on Inverse iterations http://en.wikipedia.org/wiki/Inverse_iteration . If you are looking for a high performance algorithm than further information on your problem would be needed. For instance, if your matrix is positive definite, then I would recommend one of the conjugate gradient based methods such as LOBPCG http://en.wikipedia.org/wiki/LOBPCG . A very comprehensive discussions of the state of the art methods for finding eigenvalues can be found on the web in the book Templates for the Solution of Algebraic Eigenvalue Problems: a Practical Guide http://web.eecs.utk.edu/~dongarra/etemplates/index.html.

Best regards.

I use The power method to find largest eigen values but I feels difficulty when I have to use the same method to find the smallest value.....And I think its a good method........

The power method finds the eigenvalue of a matrix A with the "largest modulus", which might be different from the largest eigenvalue of A.

The smallest eigenvalue by modulus of A is the largest of A^{-1}

I have not found that book, so I ignore what Dr Ahmed theorizes, and it is still not clear to me if your are talking about the actual value of the eigenvalues or their modulus. Assuming that you are referring to the eigenvalues of smallest and largest moduli, the answer to your question is: yes, we can find either the eigenvalue of largest or smallest modulus, without finding the other one, by means of an iterative method. As far as there is no more than one different eigenvalue of largest/smallest modulus, we find the eigenvalue of largest modulus by the power method, and the eigenvalue of smallest modulus by the inverse power method.

What I am doing now is,Ist of all take the inverse of the given matrix and then use "The Power Method" to find the "Dominant Eigen Value" of that new matrix.Now this dominant eigen value is the smallest eigen value of the given matrix.And now I will say NO to Doctor FAIZ theory(Can we find the smallest eigen value without finding the largest one).But now another question arises in my mind......"What should be the difference between dominant eigen value and the largest eigen value?"Are they same?Can we say largest value is the dominant value.

They are the same if all the eigenvalues are positive. Otherwise, it depends on the values of the minimum and maximum of the eigenvalues.

Dear Shakeel, as eigen values for a matrix may be a complex value(real part + img part), & we can not compare (in term of large/small) complex values, so it is not suitable to say that which eigen value is largest/smallest, but we can compare eigen values in term of magnitude, real part etc. so please make clear wht do mean by smallest/largest of eigen value..

The really good answer is: find all eigenvalues and see which one is smallest. Why I insert my two cents? This is because the power method is fine only when the largest (complex) eigenvalue is single, that is it is isolated from all others. Even real matrices happen to have complex eigenvalues.

You can use any smbolic computation prgramm, like mathematica to find the smallest and larger ones directly for any positive definte matrix

If you are a scientist/engineer you probably have a large matrix from a real problem.

You probably want the largest and smallest eigenvalues to get a condition number and bound errors.

If this is the case, then ignore the mathematicians as much as you can.

Don't find ALL the eigenvalues. It is very expensive.

Don't use a symbolic package. It is even slower.

For the largest eigenvalue start with a random unit vector, v.

iterate on w = Av, v = w/|w| (so v is a unit vector)

|w| converges to the largest eigenvalue quickly.

For the smallest. Replace A with A_inverse.

Iterate on: solve Aw = v for w, v = w/|w|

Smallest converges quickly to 1/|w| .

If you want to find the eigenvalue of A closest to an approximate value e_0, you can use inverse iteration for (e_0 -A)., ie. the power method of its inverse. This can also be used in the case of complex eigenvalues.

A good approximation to the lowest eigenvalue of a Hermitean matrix H may be obtained by minimizing the expectation value of H, e.g.

< v | H | v> in some linear space of normalized vectors v (Dirac notation used here). This is a consequence of the variational principle. Of course, real-symmetric matrices are also Hermitean. Such a minimization procedure also helps to decide whether the corresponding eigenvalue is degenerate.

This work also sometimes for complex-symmetric matrices.

Related are the recursion method of Haydock and the Lanczos algorithm to find the lowest few eigenvalues of large Hamiltonian matrices. Cp. e.g. http://faculty.kirkwood.edu/asoemad/citepapers/mcfarlane.pdf

Another way is, find the largest eigen value in magnitude lambada for the matrix A. Now obtain new matrix B = A - lambada . A. Now find the largest eigen- value mu of B by Powers method. Then the smallest eigen-value of the given matrix A will be mu - lambada.

@ R. Mittal: Could you prove your statement if A = diag(1,2,3)?

Yes. Here lamada =3.

B= A - lamada . I .

B = diag(-2,-1,0). So mu will be -2 (largest in magnitude of B). So smallest eigen value of A is -2 + 3 = 1.

Sorry for earlier statement . It shuould be mu + lamada.

Good. Then try now with A=-diag(1,2,3).

lamada = -3. B = A - lamada. I = diag(2, 1,0). mu = 2. So smallest eigen-value in modulus is 2 +(-3) = -1.

Sorry, I had missed the correction mu + lambda. However, for A = diag(-2,0,1) then mu + lambda = 1, which is neither the smallest eigenvalue of A, nor the eigenvalue of A with smallest modulus. In general, this method finds the eigenvalue of A of smallest modulus, possibly zero, if all the eigenvalues of A are of the same sign.

If what you're interested in is a measure of the condition of the matrix, you can use Gauss Elimination with column pivoting (you always have to use row pivoting) and simply divide the largest by the smallest pivot value. GE with column pivoting is very easy to implement. Also, there are several good methods for singular value decomposition of a matrix, some which work with non-symmetric and non-positive-definite matrices. I always use the same subroutine that I got years ago from the HP SSP. It uses Householder elimination. There's also one in IMSL that I think uses Givens rotations and one in the IBM SSP, but I can't remember what method it uses.

Using Mathematica:

Find the eigenvalues of a matrix A; as Evals.

Evals = Eigenvalues[A];

Then

Min[Evals]

The Min[Evals] will give you the minimum eigenvalue of A. Similarly use Max[Evals] for the maximum value of A.

numitr=100;

y=ones(n,1);

for k = 1 : numitr

iter = k;

v = y/norm(y,2);

y = M1\v;

th =v'*y;

if norm(y-th.*v,2) < ep*abs(th)

break;

end;

end;

V1 = y/th;