# 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 $S$?

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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.

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Schmuel Friedland proves in "Matrices with prescribed off-diagonal elements" ( http://link.springer.com/article/10.1007%2FBF02762620 ) 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.

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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.

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

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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
While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. – tomasz Aug 18 '12 at 19:19