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Let $A= a(i,j)$ and $B= b(i,j)$ be ($n\times n$) matrices that are positive definite such that $a(i,j) < b(i,j)$. Let $C= c(i,j)= b(i,j) - a(i,j)$, then $C$ is also positive definite. Why or why not?

thehandoftheking
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3 Answers3

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Dated, but was interested in this, so thought I'd provide another way of looking at it w/ Gershgorin's theorem: each eigenvalue of any square matrix $A$ lies within one of its Gershgorin disks $(A_{ii}, R_i)$, where $R_i = \sum_{j \neq i} |A_{ij}|$.

Thus if $C = B - A$, each disc is centered at $C_{ii} = B_{ii} - A_{ii} > 0$. The eigenvalues of $C$ lie in the disks $(b_{ii} - a_{ii},\, \sum_{j\neq i} |a_{ij} - b_{ij}|)$.

In the case in which $A$ and $B$ are positive semidefinite and $C_{ij} = B_{ij} - A_{ij} \geq 0$, a sufficient condition for $C$ to be positive definite then is that $C_{ii} \geq \sum_{j\neq i} (B_{ij} - A_{ij})$.

Adam
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Hint: Let $A$ be the $2\times2$ identity matrix, and let $B = A + \pmatrix{ 1 & 1 \\ 1 & 1 }$.

fuglede
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    Or maybe better, $A + \pmatrix{ 1/2 & 1 \cr 1 & 1/2 \cr}$. – Robert Israel Jun 24 '13 at 18:26
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    From this example it looks like the answer is "no." Here $A = \pmatrix{1 & 0 \ 0 & 1}$ and $B = \pmatrix{2 & 1 \ 1 & 2}$, both of which are positive definite. The difference $C = B - A = \pmatrix{1 & 1 \ 1 & 1}$ is not invertible and therefore is not positive definite. At least that's the way it looks to me. Any errors in this? Thanks, by the way. – thehandoftheking Jun 24 '13 at 18:46
  • Nope, you got it. – fuglede Jun 24 '13 at 18:48
  • @thehandoftheking Note that, in Robert Israel's example, $B-A$ is indefinite. Consequently, $A-B$ is also indefinite. So, your claim that "$a(i,j)−b(i,j)$ is negative definite" is wrong. – user1551 Jun 24 '13 at 19:39
  • @user1551 I removed that claim. – thehandoftheking Jun 26 '13 at 02:09
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In general the answer is no. As fuglede has done, we can give an explicit counter-example.

But it is also interesting to know how to tackle these problem.

The assumption is that $a(i,j)<b(i,j)$ for all $i,j$. Hence may have $b(i,j)=\varepsilon+a(i,j)$ for some coordinates $i,j$, and $b(i,j)=1+a(i,j)$ for others.

We have to take the $c(i,j)$ large at corners of the matrix and small at the diagonal, because positive definiteness of a matrix can be thought as a "dominance of the diagonal terms with respect to the non-diagonal ones".

Community
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Davide Giraudo
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