Consider the matrix $2\times2$ symmetric matrix: $$ A_2=\begin{pmatrix} 1 & a_1 \\ a_1 & 1\end{pmatrix}. $$ It's clear that the restriction $|a_1|<1$ implies that $\det(A_2)>0$. Moreover, this is the best restriction on the modulus of $a_1$ with this property, for $\det\begin{pmatrix} 1 & 1\\ 1 & 1\end{pmatrix}=0$.

Now, consider the $3\times3$ symmetric matrix: $$ A_3= \begin{pmatrix} 1 & a_1 & a_2\\ a_1 & 1 & a_3 \\ a_2 & a_3 & 1\end{pmatrix} $$ We have $\det(A_3)=1+2a_1a_2a_3-a_1^2-a_2^2-a_3^2$. Then, the restriction $|a_i|<1/2$ implies that $$ \det(A_3)>1-2(1/8)-3(1/4)=0. $$ Moreover, this is the best restriction on the modulus of the $a_i$'s with this property, for $$ \det\begin{pmatrix} 1 & -1/2 & -1/2\\ -1/2 & 1 & -1/2 \\ -1/2 & -1/2 & 1\end{pmatrix}=0. $$

Now we consider the general case. Let $A_n=[a_{i,j}]_{1\leq 1,j\leq n}$ be a matrix such that $a_{i,j}=a_{j,i}$ for every $1\leq i,j\leq n$ and $a_{i,i}=1$ for every $1\leq i\leq n$. What is the greatest number $\alpha_n$ satisfying $$ \max_{i\neq j}|a_{i,j}| < \alpha_n\ \ \Rightarrow\ \ \det A_n>0? $$

I was able to prove that $\alpha_n\leq 1/(n-1)$, by showing that the determinant of the following $n\times n$ matrix is $0$: $$ M_n=\begin{pmatrix} 1 & -\dfrac{1}{n-1} & -\dfrac{1}{n-1} & \dots & -\dfrac{1}{n-1} \\ -\dfrac{1}{n-1} & 1 & -\dfrac{1}{n-1} & \cdots & -\dfrac{1}{n-1} \\ -\dfrac{1}{n-1} &-\dfrac{1}{n-1}& 1 &\dots & -\dfrac{1}{n-1} \\ \vdots & \vdots & \vdots &\ddots & \vdots \\ -\dfrac{1}{n-1} & -\dfrac{1}{n-1}& -\dfrac{1}{n-1}&\dots & 1\end{pmatrix}. $$

One way to see that $\det(M_n)=0$ is to pick $v_1,...,v_n\in\mathbb R^n$ such that $\|v_i\|=1$ for every $i\in\{1,...,n\}$ and $\{v_1,...,v_n\}$ is the set of vertices of a regular $(n-1)$-symplex. One may se by induction that $\langle v_i, v_j \rangle=-1/(n-1)$ for every $i\neq j$. Therefore, $$ M_n=\left[\langle v_i, v_j \rangle\right]_{1\leq i, j \leq n}. $$ Since $v_1,...,v_n$ are the vertices of an $(n-1)$-symplex, they are linearly dependent, so $$ \det\left(\left[\langle v_i, v_j \rangle\right]_{1\leq i, j \leq n}\right)=0. $$

My intuition says that $\alpha_n=1/(n-1)$. However, I'm not able to prove that $\alpha_n \geq 1/(n-1)$. Any suggestions would be very helpful. Thanks in advice.

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    $\begingroup$ It's easier to note that the all ones vector is an eigenvector of $M_{n}$ with eigenvalue $0.$ $\endgroup$ Jun 4, 2017 at 9:03

1 Answer 1


Your guess is correct. If the elements outside the diagonal have absolute values less than $1/(n-1)$, the matrix has 'diagonal dominance', thus it is nonsingular.

To make the answer self-contained, I give a proof. If $x=(x_1,\dots,x_n)^t$ satisfies $Ax=0$, take $k$ such that $|x_k|$ is maximal and look at $\sum a_{ki}x_i$. The summand $a_{kk}x_k$ has greater absolute value than all other summands altogether, thus the total sum is non-zero. A contradiction.

So, determinant of $A$ is non-zero. And it is non-zero if we multiply all elements outside diagonal by $t\in [0,1]$, call such a new matrix $A(t)$. Since $f(t)=\det A(t)$ is continuous, does not vanish on $[0,1]$ and $f(0)=1$, we get $f(1)>0$ as desired.

  • $\begingroup$ I'm sorry I can't follow the argument. Could you be more specific? $\endgroup$ Jun 4, 2017 at 7:26
  • $\begingroup$ Ok, I did some research on the term diagonal dominance, then I found out the concept of strictly diagonal dominant matrix and the Levy–Desplanques theorem. Thank you very much. $\endgroup$ Jun 4, 2017 at 7:35
  • $\begingroup$ @AndréPorto , can you say something about these concepts in your OP or in a new answer? It seems interesting. $\endgroup$
    – Amir Sagiv
    Jun 4, 2017 at 7:39

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