# Can one estimate the distribution of eigenvalues of a matrix by its Cauchy/Stieltje transform?

Given a real symmetric $n$ dimensional matrix $A$, with eigenvalues $\lambda_i$ I am defining its Cauchy transform as the function, $f_A(z) = \sum_i \frac{1}{z-\lambda_i}\,$

• Is there any information about the $\lambda_i (A)$s that can be inferred from $f_A(z)$?

For example I would want an analysis of the following kind :

Say one can estimate the value of $z (A,\epsilon) \gt \max \{ \lambda_i \}$ such that for all $z \geq z (A,\epsilon)$, $f_A(z) \leq \epsilon$. Now suppose one is given a vector $v$ of the same dimension as $A$.

• Is there any connection between $\max \{\lambda(A + vv^T) \} - \max\{ \lambda(A) \}$ and $z(A + vv^T,\epsilon) - z(A,\epsilon)$? (at least for any restricted class of $A$ and $v$?)

Here I picked on $z(A,\epsilon)$ as an example of a property of the Cauchy transform which I feel might be sensitive to how the maximum eigenvalue of $A$ changes under rank-$1$ updates to $A$. But may be this $z(A,\epsilon)$ is not the right quantity - but there is may be some other property of the Cauchy transform which knows about, $\max \{\lambda(A + vv^T) \} - \max\{ \lambda(A) \}$ ?

If necessary assume that $v = (e_i \pm e_j)$ for some $j > n/2$ and $i<= n/2$.

If necessary assume that $A$ is constructed as follows : First take the matrix $D - Ad$ where $D$ is the diagonal matrix of degrees of some bi-partite graph and $Ad$ is its adjacency matrix. Then flip some of the off-diagonal $-1$ entries of $D-A$ to $1$ keeping the entire thing symmetric.

Some possibly related things I saw while searching,

Since $A$ is $n\times n$ a hermitian matrix, and thus all the eigenvalues are real, it might be of use to think of $A$ (for $n$ sufficiently large) inducing a probability measure $\mu_A(d\lambda)$. In this sense (and with some more technical poking) we can approximate the Cauchy transform in the original post with the function $$F_A(z) = \int_\mathbb{R} \frac{1}{z - \lambda}\mu_A(d\lambda)$$ This is simply the Stieltjes transform of the measure $\mu_A(d\lambda)$, and you can use the Stieltjes-Perron inversion formula to get $\mu_A$ back.

This is very useful since $F_A$ is analytic on the upper half-plane and goes to zero as $|z| \rightarrow \infty$. We also have the bound $|F_A| \le \frac{\Im{F_A}}{\Im{z}}$. There are a bunch of other results floating around on functions like this, but i have yet to find a comprehensive list.

• Thanks! What in that Mirollo link you think corresponds to my question? Jun 1 '15 at 3:07
• It handles some of the convergence issues. Jun 1 '15 at 4:48

Your family $(\lambda_i)_{i\in I}$ must reflect all the eigenvalues with their multiplicities, this means, for each $\lambda$ in the spectrum of $A$, the number of times it is repeated in the family $$n(\lambda)=card(\{i\in I|\lambda_i=\lambda\})$$ must be its multiplicity (i.e. the multiplicity of $\lambda$ as root of the polynomial $det(\lambda I-A)$)

Then $$f_A(z)=tr[(zI-A)^{-1}]$$ (this embraces the case when $k\not=\mathbb{C}$)

and you can recover the spectrum (and its multiplicities) from $f_A$ (which is rational) by decomposing it in partial fractions, precisely $$parfrac(f_A,z,\lambda)=\frac{n(\lambda)}{z-\lambda}$$ Hope it helps.

• Can you clarify what you mean by "family reflect all the eigenvalues with multiplicities". What I want to estimate is the difference between the largest eigenvalues before and after the rank-1 update as explained. Does the Cauchy transform or some variant of it help see that? Jun 4 '15 at 23:41
• Yes, when one one states "with eigenvalues $\lambda_i$" one means that the family $(\lambda_i)_{_\in I}$ is without multiplicities and sometimes, repeated as much as the multiplicity is. For example, with the diagonal matrix $(2,2,1)$, in the first case, the family can be $(1,2)$ and $(2,1,2)$ in the second. I clarify it in the text. Jun 5 '15 at 3:27
• I suppose your maximum $\max \{\lambda(A + vv^T) \}$ can be taken over some vectors of a certain kind (maybe not only $\pm$ basic vectors) ? Jun 5 '15 at 3:51
• Well - my A and v are restricted to the kinds that I mentioned in the post. The "max" is the maximum eigenvalue of this matrix. Lets say I am given a $A$ and a $v$ of the types mentioned. Then I want to say something about how much the max eigenvalue of $A + vv^T$ is ahead if the max eigenvalue of $A$ Jun 6 '15 at 18:54
• @Anirbit Thank you for the explanation. If I have idea(s) that can help, I'll extend my answer. Jun 7 '15 at 8:14