Show that a matrix has positive determinant

Question

Let $A$ be an $n\times n$ matrix, where $a_{ii}>0$ and $a_{ij}\le 0$ for $1\le i\ne j\le n$ and also $\sum_{i = 1}^n a_{ij}>0$, show that $\det(A)>0$.

I try to use the fact that $$\left(\sum_{i = 1}^n a_{i1}e_i\right)\wedge\cdots\wedge \left(\sum_{i = 1}^n a_{in}e_i\right)= A_1\wedge \cdots \wedge A_n = Ae_1\wedge \cdots\wedge Ae_n $$

$$= A\wedge \cdots \wedge A(e_1\wedge\cdots\wedge e_n) = \det(A)e_1\wedge\cdots\wedge e_n$$

but not sure how to proceed.

Answer 1

It's not hard to see that $a_{jj}>\sum_{i=1, i \neq j} |a_{ij}|$, that is, $A$ is strictly diagonally dominant (actually $A^t$ is strictly diagonally dominant, but there is no problem here since $\det{A}=\det{A^t}$.

So, using that:

• the eigenvalues of strictly dominant matrices have positive real part if all diagonal entries are positive (see Theorem 6.1.10 of Horn and Johnson, for example);
• if $z \in \mathbb{C}$ is an eigenvalue of $A$, then its conjugate $\bar{z}$ is also an eigenvalue, since the characteristical polynomial of $A$ has real coefficients;
• $\det{A}$ is the product of the eigenvalues of $A$.

the result proceeds easily.

Answer 2

So here is the proof (it isn't mine, I remembered that I had seen it one of my textbooks and I was right). The proof assumes that your inequality true for rows rather than for columns, but it doesn't change things much. It is easy to see that $$|a_{ii}|\gt\sum_{j=1,j\ne i}^{n}|a_{ij}|,~~~i=1,2,\ldots,n:$$ Since $a_{11}\ne 0$ we can use Gauss elimination method. After doing that we have $$ A^{(1)}=\begin{pmatrix} 1&u_{12}&\ldots&u_{1n}\\ 0&a_{22}^{(1)}&\ldots&a_{2n}^{(1)}\\ \ldots&\ldots&\ldots&\ldots\\ 0&a_{n2}^{(1)}&\ldots&a_{nn}^{(1)}\\ \end{pmatrix},$$ and it's easy to see that $$\sum_{j=2}^{n}|u_{1j}|=\sum_{j=2}^{n}\frac{|a_{1j}|}{|a_{11}|}\lt1:$$ Now if we prove that $$|a_{ii}^{(1)}|\gt\sum_{j=2,j\ne i}^{n}|a_{ij}^{(1)}|,~~~i=2,\ldots,n:$$ we will complete the proof. Let's do that. $$ |a_{ii}^{(1)}|=|a_{ii}-a_{i1}u_{1i}|\ge|a_{ii}|-|a_{i1}||u_{1i}|\gt\sum_{j=1,j\ne i}^{n}|a_{ij}|-|a_{i1}||u_{1i}|=\sum_{j=2,j\ne i}^{n}|a_{ij}|+|a_{i1}|-|a_{i1}||u_{1i}|=\sum_{j=2,j\ne i}^{n}|a_{ij}^{(1)}+a_{i1}u_{1j}|+|a_{i1}|-|a_{i1}||u_{1i}|\ge\sum_{j=2,j\ne i}^{n}|a_{ij}^{(1)}|-\sum_{j=2,j\ne i}^{n}|a_{i1}u_{1j}|+|a_{i1}|-|a_{i1}||u_{1i}|=\sum_{j=2,j\ne i}^{n}|a_{ij}^{(1)}|+|a_{i1}|-\sum_{j=2}^{n}|a_{i1}||u_{1j}|=\sum_{j=2,j\ne i}^{n}|a_{ij}^{(1)}|+|a_{i1}|\left(1-\sum_{j=2}^{n}|u_{1j}|\right): $$ I hope I didn't do any mistake:)