Davis-Kahan Theorem

[[concept]]

Theorem (Davis-Kahan)

Let $T$ and $T^{\prime }$ be self-adjoint Hilbert-Schmidt operators with eigenspectra (${\lambda }_{i,},{\phi }_i$) and $({\lambda }_i^{\prime },{\phi }_i^{\prime })$ respectively ordered by eigenvalue magnitude. Then $||{\phi }_i-{\phi }_i^{\prime }||\le \frac{\pi }{2}\frac{||T-T^{\prime }||}{d({\lambda }_i,{\lambda }_i^{\prime })}$ Where $||\cdot ||$ is the operator norm and $d({\lambda }_i,{\lambda }_i^{\prime })$ is defined as $d({\lambda }_i,{\lambda }_i^{\prime })\equiv \min ({\min }_{j\ne i}(|{\lambda }_i-{\lambda }_j^{\prime }|),{\min }_{j\ne i}(|{\lambda }_j-{\lambda }_i^{\prime }|))$ Call the first interior min $d_1$ and the second $d_2$

Example

$d({\lambda }_i,{\lambda }_i^{\prime })$ corresponds to the minimum “eigengap” for the closest eigenvalue for eigenvalue ${\lambda }_i$ in its own spectrum.

NOTE

If $d(\cdot ,\cdot )$ is large, then this is OK for fast convergence. But if this is small, then we also need small $||T-T^{\prime }||$ to get fast convergence.

• Process into its own note? edit

Let $\{G_n,X_n\}\to (W,\mathcal{X})$. We can use the Davis-Kahan Theorem with $T_W$ (the limiting graphon shift operator) and $T_{W_n}$ (induced graphon shift operator). $||\phi (T_{W_n})-{\phi }_i(T_W)||\le \frac{\pi }{2}\frac{||T_W-T_{W_n}||}{d({\lambda }_i(T_W),{\lambda }_i(T_{W_n}))}$ This is good because from lecture 15, we know that if $G_n$ converges to $W$ in the cut norm, then $T_{W_n}$ converges to $T_W$ in $L_2$ (cut norm Convergence in the cut norm implies convergence in L2). Thus we have convergence in the numerator!

Does this mean everything is OK?

• No! The denominator might vanish as $i\to \infty$ since we have ${\lambda }_i\to 0$ from the second part of the spectral theorem for self-adjoint compact operators on Hilbert spaces
  • even though we look at $\min (d_1,d_2)$ and ${\lambda }_i\ne {\lambda }_j$ converge to different values (and therefore $d(\cdot )\to \ell >0$ ), it still might not work.
• Even if we have eigenvalue convergence, this convergence is not necessarily uniform. (recall the difference between convergence and uniform convergence) because the graphon eigenvalues accumulate at 0.

Mentions

Mentions

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