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I have a Hermitian matrix $\mathbf{H}$ which depends on two parameters say $x$ and $y$. When I diagonalize it at two close points $(x_1,y_1)$ and $(x_2,y_2)$ I get two close eigenvalues ($\varepsilon_1$ and $\varepsilon_2$) and two corresponding eigenspaces ($S_1$ and $S_2$) of the same dimension.

Note that they are not eigenvalues of the same matrix. There are two different matrices: $\mathbf{H}_1=\mathbf{H}(x_1,y_1)$ and $\mathbf{H}_2=\mathbf{H}(x_2,y_2)$.

I have a mesh of points $(x_i,y_i)$ and want to find the eigenvalue and the eigenspace at any point using interpolation. The problem is that since the matrices are diagonalized numerically the bases of $S_1$ and $S_2$ are completely independent. Even if $(x_1,y_1)$ and $(x_2,y_2)$ are very close the basis vectors can have very different components.

For interpolation I need a basis that depends on $x$ and $y$ continuously, i.e. the closer the eigenspaces $S_1$ and $S_2$ the closer should be the basis vectors.

If $S_1$ and $S_2$ are plains in 3-dimensional Euclidean space then a good way to select a basis in S2 is to rotate the basis of S1 around the line which is the intersection of the plains. Is there something analogous to this in complex multidimensional space?

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For simplicity, assume that there is only one parameter $t$ rather than your two.

In order that you can have continuous eigenspaces, you need to assume that the associated eigenvalues do not nearly cross. (For nearly crossing eigenvalues it may very well happen that the eigenspaces are essentially exchanged though the eigenvalue curves do not touch. This happens regularly in symmetric parametric eigenvalue problems and is called the phenomenon of avoided crossings.)

If the eigenspace of $H(t)$ to the continuous eigenvalue $\lambda(t)$ is the column space of $Q_k$ when $t=t_k$, you need to assume a form $Q(t)=Q_0+\sum_j \Phi_j(t) Z_j$ with fixed basis functions $\Phi_j(t)$ (e.g., B-splines) and $Q_0$ being one of the $Q_k$, and then fit the coefficient matrices $Z_j$ and (low-dimensional eigenspace transforms $Y_k$ to the approximate equation $Q_k Y_k - \sum_j \Phi_j(t_k) Z_j \approx Q_0$. This is a linear least squares problem that is easy to solve.

With 2 parameters, replace the B-splines by FEM shape functions. Your least squares problem might now become large and suitable additional trickery is needed to make the problem solvable if the direct solution is infeasible.

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  • $\begingroup$ Thank you for the answer! Not everything is clear for me though. 1) Does $\Phi$ depend on $j$? If not, why there is a sum? 2) Is $Y_k$ a linear transform inside the eigenspace $Q_k$? $\endgroup$ – Maksim Zholudev Apr 30 '12 at 10:53
  • $\begingroup$ I corrected the missing index of the basis functions. Yes, $Y_k$ changes the basis inside the $k$th eigenspace. $\endgroup$ – Arnold Neumaier Apr 30 '12 at 11:19

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