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$A=(a_{ij})$ 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 Answers 1

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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 $1$.

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    if i am not mistaken, gershgorin gives you only $0 < \lambda < 2$ – 2017-01-11
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    yeah, then use the trace and AM-GM to finish. – 2017-01-11
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    This is a nice solution, but I would swear there's a much more elementary way, as well... perhaps just fault memory? – 2017-01-11
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    Sure! I didn't know that theorem. I think it solves the problem. – 2017-01-11
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    It's a pretty cool thoerem, I don't know why most courses don't teach it though. – 2017-01-11
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    @arnaldo, but this shows that every eigenvalues is inside the interval $(1-1,1+1)$. So it would still be possible for the determinant to be greater than $1$. – 2017-01-11
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    Oh, right! I need more coffee! Very nice solution. Thanks for present me the Gershgorin theorem. – 2017-01-11
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    thanks for the explanation. i did not see the am-gm. nice. – 2017-01-11
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    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. – 2017-01-11