Let $A \in \mathbb{R}^{n \times n}$ be a symmetric matrix with entries $A_{ij} \sim \mathcal{N} (0,1)$, all independent except for the symmetry condition.

Consider the following minimization problem:

\begin{equation} \inf \limits_{u \in \mathbb{R}^n : \sum u_i = 0 }\{ \ \lambda_{max}( A - \text{Diag}(u))\} \end{equation}

where $\lambda_{max}(\cdot)$ denotes the largest eigenvalue and $\text{D}(v)$ is the diagonal matrix having the vector $v \in \mathbb{R}^n$ as entries.

I don't want to necessarily find the optimal value of this problem. It would suffice to find a vector $u$ that achieves a smaller value than the trivial vector $0 \in \mathbb{R}^n$, for large $n$.

Question: How would one find a vector $u$, with $\sum_i u_i = 0 $ (possibly random and dependent on $A$), that achives a smaller value than the zero vector asymptotically for large $n$ with high probability? For the zero vector the value is $\lambda_{max}(A)$ for which it is know that $\lambda_{max}(A) = \Theta( 2 n^{3/2}) $.

Is there existing literature on this problem? What methods might one use?

  • $\begingroup$ Out of curiosity, why do you want that $\sum u_i = 0$? $\endgroup$
    – smapers
    Feb 5, 2020 at 15:50
  • $\begingroup$ The mentioned problem can be viewed as an upper bound to the problem of maximizing $x^t A x$ over the hypercube. $\endgroup$
    – sigmatau
    Feb 5, 2020 at 15:59
  • $\begingroup$ I would like to find a vector such that the value is smaller than the value attained by the zero vector with high probability for $n \to \infty$. $\endgroup$
    – sigmatau
    Feb 5, 2020 at 16:05
  • $\begingroup$ I think that due to the $\Sigma u_i = 0$ constraint and the convexity of $\lambda_{max}$, Jensen's inequality is going against you. $\endgroup$ Feb 5, 2020 at 16:38
  • 1
    $\begingroup$ @smapers That was not clearly written out. The expectation is (sort of?) conditional on the value of A. I.e., randomly reordering elements of u for a fixed $A$. Anyhow, this is not a rigorous argument, rather, a germ of an idea, and perhaps it is wrong. That is why I did not submit it as an answer. $\endgroup$ Feb 6, 2020 at 13:23

1 Answer 1


For a given instantiation of $A$, the globally optimal value of $u$ can be found as the solution of a convex Linear Semidefinite Programming (SDP) problem using a standard solver such as Mosek. This optimum achieves a better objective value than $u = 0$ a.s. for $n \ge 2$.

Now assume, I believe as intended, that $u$ must be specified only with knowledge of $n$, but not with knowledge of a specific instantiation of $A$.

By convexity of $\lambda_{max}$ of a real symmetric matrix, $$\lambda_{max}(A) \le \frac{1}{2}\lambda_{max}(A-\text{diag}(u)) + \frac{1}{2}\lambda_{max}(A-\text{diag}(-u))$$

Taking expectation of both sides, $$E(\lambda_{max}(A)) \le \frac{1}{2}E(\lambda_{max}(A-\text{diag}(u))) + \frac{1}{2}E(\lambda_{max}(A-\text{diag}(-u)))$$

Because $u$ was chosen without knowledge of the instantiation of $A$, and all elements of $A$ are symmetrically distributed about zero, it must be the case that $$E(\lambda_{max}(A-\text{diag}(-u))) =E(\lambda_{max}(A-\text{diag}(u)))$$

and therefore $$E(\lambda_{max}(A)) \le E(\lambda_{max}(A-\text{diag}(u)))$$

So such a $u$ can do no better on average than $u = 0$.

  • $\begingroup$ Does the downvoter care to explain the downvote? $\endgroup$ Apr 21, 2020 at 1:46

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