I have a set of matrix, which is:

  1. Real symmetric positive definite. Very sparse.
  2. Diagonal elements are positive while off-diagonal elements are negative.
  3. $\displaystyle a_{ii}=-\sum^{n}_{{j=1}\atop{j\ne i}} a_{ij}$
  4. $a_{ii} \in (0,1]$
  5. $a_{ij} \in (-1,0]$ when $ i \neq j$

My experiments show that the largest eigenvalue of all the matrices I have are larger than 1. Can some one help me on proving that $ \lambda_{max} >1 $ for this matrix?

My first though is to prove that $Ax=\lambda x < x$ does not hold. But I couldn't get any break through. Thanks!


Your conjecture can't be true unless you put some constraints on the diagonal entries or something. If you have a matrix $A$ that has eigenvalue $\lambda$, then the matrix $kA$ has eigenvalue $k\lambda$ with the same eigenvector space. And your matrix conditions are invariant under multiplying the matrix by a positive constant. So given a matrix that satisfies your constraints, you can find an arbitrarily small matrix with arbitrarily small eigenvalues that satisfies the same constraints.

  • $\begingroup$ If there is always a '1' element in the diagonal element, can it be counted as the constraints you mentioned? Therefore the row the '1' appeared will be like: $x_{i} + \sum^{n}_{j=1} a_{ij}x_{j} < x_{i}$ for $Ax=\lambda_{max}x<x$. Thanks! $\endgroup$
    – Jane Li
    May 1 '14 at 17:01
  • $\begingroup$ @JaneLi: Your last statement does not make sense. When there is a $1$ on the diagonal, the maximal eigenvalue is no less than $1$ since $\lambda_{\max}=\max_{\|x\|_2=1}x^TAx$. Pick an $x$ to be the vector with $1$ on the same row as $1$ appearing on the diagonal of $A$ and zero elsewhere. $\endgroup$
    – Hans
    Dec 28 '17 at 18:54

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