Sensitive eigenvectors to small perturbations in the matrix?

Question

I've encountered a mathematical issue in my research. To provide some context, I have a known density matrix that I am reconstructing numerically using quantum data. The rebuilt matrix has fluctuations but it is very close to the original one.

Now, as I attempt to find the eigenvectors, I expect both the eigenvalues and eigenvectors to closely resemble the original ones. Surprisingly, it seems true for the eigenvalues and two of the three eigenvectors. However, the third eigenvector consistently turns out quite different from the original, even when the perturbations are minimal.

Here is one out of many examples :

$ \mathbf{\rho} $ matrix:

\begin{pmatrix} 0.25 & -0.0625 - 0.1875i & -0.1875 + 0.0991i \\ -0.0625 + 0.1875i & 0.25 & -0.1875 - 0.0991i \\ -0.1875 - 0.0991i & -0.1875 + 0.0991i & 0.5 \end{pmatrix}

$\lambda_2 \approx 0, \quad \vec{e}_2 = (0.46697442 + 0.46697442i, \quad 0.66040155, \quad 0.33020078 + 0.13677364i) $

$ \rho_{num} $ matrix: \begin{pmatrix} 0.2496 & -0.0711 + 0.1779i & -0.1762 - 0.1063i \\ -0.0711 - 0.1779i & 0.252 & -0.1816 + 0.0998i \\ -0.1762 + 0.1063i & -0.1816 - 0.0998i & 0.4984 \end{pmatrix}

$\lambda_2 \approx .009, \quad \vec{e}_2 = (0.66282818 , \quad 0.49280465 + 0.43932726i , \quad 0.33239702 + 0.11960834i) $

On the first sight, it looks like somehow the first and the second entries of the ion vectors are swapped. But I think the reason this is happening is not because of the swap; it's actually a change of phase. It is still the same number but with a different phase. In any case, I'm not sure why this is happening.

Not sure the error's in my approach or if it's inherent to the nature of the problem. Can you please share your insights, and where should I look to fix it? Thank you.

Answer 1

This has nothing to do with perturbations. Your approximation matrix is conjugated with regards to the original, so you would expect your eigenvector to be close to the conjugate of the original $\lambda_2$.

Your eigenvector that you found is indeed very close, but it is just multiplied by a scalar. If you multiply $\lambda_2$ from the $\rho_{num}$ by $\frac{0.33239702-0.11960834i}{0.33239702+0.11960834i}$, you get (rounded) $$ (0.5109-0.4223i, 0.6597+0.0246i, 0.3324-0.1196i) $$ And that is pretty close to the value you would expect it to be.

Answer 2

This is hardly surprising, as the new matrix is not close to the old one. In fact, the relative error $\|\rho_{num}-\rho\|/\|\rho\|=0.6176$ between the two matrices is very large.

What happens here is that $\rho_{num}$ is close not to $\rho$, but to the transpose of $\rho$. Hence $\rho_{num}$ and $\rho^\top$ have similar spectrum and similar eigenspaces. However, the eigenvectors of the Hermitian matrix $\rho^\top$ are not the eigenvectors of the Hermitian matrix $\rho$, but the complex conjugates of them. In your case, if we denote the $\vec{e}_2$ for $\rho$ by $u$ and the one for $\rho_{num}$ by $v$, then $\operatorname{span}\{v\}$ should be close to $\operatorname{span}\{\overline{u}\}$. Indeed, we have $v\approx(0.7290 + 0.6800i)\overline{u}$.