Suppose I have an (asymmetric) matrix $A$ with a known eigendecomposition. I'm trying to establish any way of bounding the difference/angle between the leading eigenvector $v_1$ and leading singular vector $w_1$.
I am sure there will be counterexamples that can show the two can be arbitrarily distinct but my question is whether any conditions exist (aside from symmetry) that can provide a bound? For example, it is clear that the distance tends to grow as the symmetry of $A$ decays.
Attempts:
For example, suppose we could find the $L_2$-closest symmetric matrix $\tilde{A}$ with the same leading eigenvector/eigenvalue. Then its singular vector equals its eigenvectors, so we can deploy Wedin's sin theorem to get:
$$ \sin(\angle v_1, w_1) \leq \frac{||A-\tilde{A}||_2}{\sigma_1-\tilde{\sigma_2}} $$
One way to find a bound would therefore be to find this closest $\tilde{A}$ and establish the values on the right hand side, but it's not clear to me exactly what this matrix will be.
- The real part ($A+A^T$)/2 would be a natural guess, but then the leading eigenvector will change.
- One could even construct a matrix $\tilde{A}$ by taking the Gram-Schmidt orthogonalisation of the eigenvector matrix of $A$ (which would retain the leading eigenvector and all eigenvalues), but then the norm would be difficult to compute in closed form
Any ideas welcome!