# Changing the spectrum of a symmetric matrix by diagonal perturbations

Given a fixed symmetric matrix $S$, can one change the spectrum of $S$ to any desired set of eigenvalues $\{\lambda_1,\dots,\lambda_n\}$ by adding a diagonal matrix $D$ to $S$?

• Unfortunately not. – Manos Oct 18 '11 at 23:28
• For a special case take a look at this – user13838 Oct 18 '11 at 23:33

Let $A$ be a given symmetric matrix over the reals and let $B$ be a symmetric matrix over the reals as well that we add to $A$ to obtain $\tilde{A}=A+B$. Let $B=S \Lambda_B S^{-1}$ be the eigendecomposition of $B$. Then $\tilde{A} = S\left(S^{-1}AS + \Lambda_B\right)S^{-1}$. From that we see that the eigenvalues of $\tilde{A}$ are precisely the eigenvalues of $S^{-1}AS + \Lambda_B$. Since the eigenvalues of $S^{-1}AS$ are identical to those of $A$, we see that, for the purpose of spectrum modification, we lose no generality if we perturb $A$ with a diagonal matrix initially.

So let's consider $\tilde{A}=A+B$, where $B$ is diagonal.

Assuming no specific structure on $A$, the answer to the question "what is the spectrum of $\tilde{A}$" is not known, to the best of my knowledge. Instead, there exist some interesting theorems that give information about the distribution of the spectrum of $\tilde{A}$ with respect to that of $A$ and $B$.

The ones more interesting to me, are two theorems by the great Hermann Weyl, known as Weyl 1 and Weyl 2. Let the eigenvalues of $A$ be ordered as $\lambda_1(A) \le \lambda_2(A) \cdots \le \lambda_n(A)$ and similarly for the other matrices. Then

(Weyl 1)

$\lambda_k(A)+\lambda_1(B) \le \lambda_k(A+B) \le \lambda_k(A) + \lambda_n(B)$

and

(Weyl 2)

$\lambda_{j+k-n}(A+B) \le \lambda_j(A) + \lambda_k(B) \le \lambda_{j+k-1}(A+B)$

where we interpret an eigenvalue corresponding to an index greater than n as plus infinity and less than 1 as minus infinity.

Schmuel Friedland proves in "Matrices with prescribed off-diagonal elements" (link) that, given M any n-by-n matrix over $\mathbb{C}$, there exists a diagonal matrix (also taken over $\mathbb{C}$ ) such that $M+D$ has exactly the eigenvalues you want.

I don't know of any related results specific to M an n-by-n real symmetric matrix.

If $S$ is symmetric and $D$ is diagonal then $S+D$ is symmetric, so its eigenvalues are all real, so if your "desired set" contains any nonreal numbers you are bound to be disappointed.

Please read the paper "Honeycombs and sums of Hermitian Matices" by A. Knutson and T. Tao

• a bit more detail about how the paper might help or where in the paper to look would be helpful. Perhaps a link to the paper would be nice. – robjohn May 16 '12 at 17:16