$A=(a_{ij})$ is a $n\times n$ symmetric real matrix such that:

$a_{ii}=1$ and $\sum_{j=1}^{n}|a_{ij}|<2$ for all $i \in \{1,2,3,...,n\}$.

Prove that $0< \det(A) \le 1$.

My approach:

That is a question that I have tried before and I am trying again but still without success.

I'm trying to use spectral theorem (maybe prove that $|\lambda| \le1$) but I got nothing.

I also tried brute force using the definition (using permutations) of $\det A$.

Any idea?


1 Answer 1


All eigenvalues are real, this combined with the gershgorin disk theorem implies every eigenvalue is in the range $(0,2)$. So the determinant is positive.

Notice that the trace of the matrix is $n$, and this is equal to the sum of the eigenvalues, so by the arithmetic-geometric mean the product of the eigenvalues is at most $1$.

  • $\begingroup$ if i am not mistaken, gershgorin gives you only $0 < \lambda < 2$ $\endgroup$
    – abel
    Jan 11, 2017 at 17:51
  • 1
    $\begingroup$ This is a nice solution, but I would swear there's a much more elementary way, as well... perhaps just fault memory? $\endgroup$
    – The Count
    Jan 11, 2017 at 17:54
  • 2
    $\begingroup$ Oh, right! I need more coffee! Very nice solution. Thanks for present me the Gershgorin theorem. $\endgroup$
    – Arnaldo
    Jan 11, 2017 at 18:02
  • 1
    $\begingroup$ thanks for the explanation. i did not see the am-gm. nice. $\endgroup$
    – abel
    Jan 11, 2017 at 18:03
  • 1
    $\begingroup$ An alternative to Gershgorin: $\|(A - I)\|_1 < 1$, where $\|\cdot\|_1$ is the induced $1$-norm. We could also use $\|\cdot\|_\infty$ in the same way. $\endgroup$ Jan 11, 2017 at 19:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .