![]() In case you have a Hermitian matrix you have to use the inner product defined for complex Hilbert spaces. In this case, the second column in V corresponds to the second smallest eigenvalue D(2,2). The smallest eigenvalue is zero, indicating that the graph has one connected component. Now you will have a negative-definite matrix with the targeted eigenvalue $x'$ having the highest magnitude which you can compute using power-method.Ħ)Solve for the variable $x$: $x'=(x-\lambda)^2+max(spec(H'))$ You don't have the information of whether your initial eigenvalue was positive or negative but this can be found by multiplying the original matrix to the eigenvector. The Fiedler vector is the eigenvector corresponding to the second smallest eigenvalue of the graph. You can also use the fact that the matrix spectrum will be bounded by the Hilbert-Schmidt norm and avoid step 1.ĥ)Shift by the maximum eigenvalue/bound $H'=H'- max(spec(H'))I$. I can find them using the inverse iteration, and I can also find the largest one using the power method. This will make your matrix positive definite.Ĥ)Now the desired eigenvalue will be as close to zero, while the change in the largest magnitude eigenvalue can be computed trivially. I need to write a program which computes the largest and the smallest (in terms of absolute value) eigenvalues using both power iteration and inverse iteration. In case you are interested in the smallest magnitude eigenvalue $\lambda = 0$.ģ)Square your matrix $H'= (H-\lambda I)^2$. Assume you can calculate the eigenvalue with maximum absolute value $\omega$ using power method.Ģ)Shift the matrix by a constant $\lambda$ to target the part of the spectrum that you are interested $H-\lambda I$. In matlab the lowest eigenvalue may be found by. Detecting the shift in sign for the lowest eigenvalue indicates the point the matrix becomes singular. A general creative answer to this question can be composed the following way:ġ)Your symmetric/hermitian Matrix $H$ has a spectrum with positive and negative eigenvalues. If at least one eigenvalue is zero the matrix is singular, and if one becomes negative and the rest is positive it is indefinite.
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