3
$\begingroup$

Suppose $\mathbf{A}(\mu)$ being a symmetric positive definite matrix of dimension $n$ where its elements depend parametrically on the real parameter $\mu$.

Suppose now to build the orthonormal basis of the Krylov subspace from an initial normalized guess $\mathbf{x}_0$ by performing $m$ iterations of the Lanczos-Arnoldi algorithm. This orthonormal basis is gathered in the unitary matrix $\mathbf{V}$, dimension ($n,m$), and transforms $\mathbf{A}$ into a tridiagonal symmetric matrix $\mathbf{T}$ of dimension $m$ which can be finally diagonalized by a unitary transformation $\mathbf{U}$ as shown below where $\boldsymbol{\Lambda}$ denotes the final diagonal matrix. \begin{equation} \begin{split} \mathbf{T}&=\mathbf{V}^T\mathbf{AV}\\ \boldsymbol{\Lambda}&=\mathbf{U}^T\mathbf{TU}\end{split} \end{equation} We are now interested in the derivative of $\boldsymbol{\Lambda}$ with respect to the parameter $\mu$: this very problem was already discussed in a previous question (Derivative of eigenvectors of an Hermitian matrix) and the result is reported below. \begin{equation} \frac{d\boldsymbol{\Lambda}}{d\mu}=\mathbf{U}^T\frac{d\mathbf{T}}{d\mu}\mathbf{U} \end{equation} In our case, however, $\mathbf{T}$ is obtained via the trasfomation $\mathbf{V}$ which is the orthonormal basis of the Krylov subspace mentioned.

The question is therefore, can I compute the derivative of $\mathbf{T}$ by simply differentiating $\mathbf{A}(\mu)$ (equation below), i.e. similarly to the case for $\boldsymbol{\Lambda}$ where we ignored the derivatives of the transformation matrix, or should I consider differentiating $\mathbf{V}$ as well? \begin{equation} \frac{d\mathbf{T}}{d\mu}=\mathbf{V}^T\frac{d\mathbf{A}}{d\mu}\mathbf{V} \end{equation}

$\endgroup$
2
  • $\begingroup$ no, in the equation for $d\Lambda/d\mu$ it is used that the matrix $\Lambda$ is diagonal, so there is no analogous equation for $dT/d\mu$. $\endgroup$ Apr 21, 2021 at 17:44
  • $\begingroup$ If we write explicitly $\boldsymbol{\Lambda}=\mathbf{U}^T\mathbf{V}^T\mathbf{A}\mathbf{VU}$, we can gather the two unitary transformations into $\mathbf{W}=\mathbf{VU}$, in this case could we then differentiate $\mathbf{\Lambda}$ as follows? \begin{equation} \frac{d\boldsymbol{\Lambda}}{d\mu}=\mathbf{W}^T\frac{d\mathbf{A}}{d\mu}\mathbf{W} \end{equation} Is this valid even in the case of $\mathbf{W}$ having dimension $(m,n)$ and thus not being the full transformation containing all the $n$ eivengectors or $\mathbf{A}$? $\endgroup$
    – wolfram
    Apr 22, 2021 at 5:58

1 Answer 1

1
$\begingroup$

The magic of "not having to differentiate the eigenvectors" is known as the Hellmann–Feynman theorem. Let me walk you through it, at the level of a single eigenvalue, to answer the question you asked in the comment.

You decompose the real symmetric matrix $A$ as $A=W\Lambda W^T$, with $W$ the orthogonal matrix of eigenvectors and $\Lambda={\rm diag}\,(\lambda_1,\lambda_2,\ldots)$ the diagonal matrix of eigenvalues. Consider one eigenvalue $\lambda_k$ and the associated eigenvector $\psi$ with elements $\psi_i=W_{ik}$.
By construction, the eigenvalue equals the inner product $$\lambda_k=(\psi , A\psi)=(A\psi , \psi),$$ because $(\psi,\psi)=1$. Now take the derivative with respect to $\mu$, denoted by a prime: $$\lambda'_k=(\psi',A\psi)+(A\psi,\psi')+(\psi,A'\psi)=$$ $$\qquad=\lambda_k(\psi',\psi)+\lambda_k(\psi,\psi')+(\psi,A'\psi)$$ $$\qquad=\lambda_k\frac{d}{d\mu}(\psi,\psi)+(\psi,A'\psi)$$ $$\qquad=0+(\psi,A'\psi).$$ So you see, the derivative of the wave functions drops out because of the normalization.

$\endgroup$
2
  • 1
    $\begingroup$ Dear Carlo, than you very much you your kind explanation. I think I get you point. So, as long as $\mathbf{W}$ contains the eigenvectors of $\mathbf{A}$, I can differentiate $\boldsymbol{\Lambda}$ without caring of the "eigenvectors' response", i.e. $\mathbf{W}$. What made me dubious was the fact that, by working in a Krylov subspace ($m<n$), the final m eigenvalues of $\boldsymbol{\Lambda}$ are only approximations of the actual eigenv. of $\mathbf{A}$, but as long as the eigenvalue problem $\mathbf{AW}=\mathbf{W}\boldsymbol{\Lambda}$ is solved, the Hellmann Feynman Th. can still be applied. $\endgroup$
    – wolfram
    Apr 22, 2021 at 7:15
  • 1
    $\begingroup$ In fact, as far as I remember from the old Quantum Chemistry courses, the Hellmann-Feynman theorem is valid both for exact and approximate wavefunctions, as long as the latter are solved variationally, and this could be transferred to the case I reported where we are approximating the exact eigenvalues of $\mathbf{A}$ in a "variational fashion" since we are solving an eigenvalue problem. $\endgroup$
    – wolfram
    Apr 22, 2021 at 7:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.