# Question from Berkley Problems in Mathematics. Show that det($A$) > 0 [duplicate]

Let $A = (a_{ij})$ be an $n\times n$ real matrix satisfying the conditions: $a_{ii} > 0$ $(1\leq i \leq n)$; $a_{ij} \leq0$ $(i\not=j,1\leq i,j \leq n)$; $\sum\limits_{i=1}^n a_{ij}>0$ $(1\leq i \leq n)$. Show that det($A$) > $0$ .

## marked as duplicate by mlc, user91500, user26857, The Dead Legend, Namaste linear-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 7 '17 at 11:17

• Hint: Show that if $\lambda$ is a real eigenvalue then it is positive. – user379195 May 7 '17 at 6:54

The matrix $A$ is diagonally dominant and so nonsingular. Replacing off-diagonal entries $a_{i,j}$ by $ta_{i,j}$ for $0\le t\le 1$ also gives a diagonally dominant and nonsingular matrix $A_t$. But $A_0$ is diagonal with positive determinant. By continuity of determinant $\det A_t>0$ and so $\det A=\det A_1>0$.

• A weakly diagonally dominant matrix may be singular. – mlc May 7 '17 at 7:00
• @mlc There's nothing weak about these matrices! – Lord Shark the Unknown May 7 '17 at 7:01
• \begin{bmatrix} 1 & -2 & -2 \\ -1 & 6 & -1 \\ -1 & -1 & 3 \end{bmatrix} what about this matrix; It's not a DD matrix. But confusingly, it is satisfying the question but not the result to be shown. I must be wrong somwhere. – Hirakjyoti Das May 7 '17 at 7:09
• @HirakjyotiDas one of your conditions is that the column sums are positive. – Lord Shark the Unknown May 7 '17 at 7:13
• That's the very thing I was missing; Thank you – Hirakjyoti Das May 7 '17 at 7:16

In addition to the other answers, one can more generally see that from the Gershgorin circle theorem

Here firstly $A$ is diagonally dominant. Secondly $A$ is $Z$-matrix as all its off diagonal entries are $\leq 0.$

Also diagonal entries are $> 0.$ Hence $A$ is a $P$-matrix. (All principal minors are positive).

So $\det(A) > 0.$