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I know why in a symmetric positive definite matrix every entry on the trace is positive entry $a_{ii}>0$. However I don't how to show that the largest value of the matrix is also on it's trace, meaning a $z$ exist with $a_{zz} = max \vert a_{ij} \vert $ for $1\le i,j\le n$.

Thanks for help.

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  • $\begingroup$ when $i \neq j,$ what do you know about $a_{ii} a_{jj} - a_{ij}^2? $ $\endgroup$ – Will Jagy May 17 '15 at 19:49
  • $\begingroup$ Hello, well it has to be positive because every minor of the matrix must have a positive determinant. $\endgroup$ – Matriz May 17 '15 at 19:55
  • $\begingroup$ and what about the larger of $a_{ii}, a_{jj}?$ $\endgroup$ – Will Jagy May 17 '15 at 19:57
  • $\begingroup$ One of $a_{ii}$,$a_{ji}$ must be the largest because it's a symmetric matrix and therefore for every 2x2 minor I have $a_{ii}*a_{jj}$ minus a square. $\endgroup$ – Matriz May 17 '15 at 20:06
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First, every diagonal element of $A$ is positive since $$ 0<e_i^\top A e_i=a_{ii} $$

Let $a_{kk}$ be the largest diagonal element of $A$ and let $i\neq j$. Since $A$ is positive-definite, every principal submatrix of $A$ is positive-definite. In particular, $$ 0<\det \begin{bmatrix} a_{ii} & a_{ij} \\ a_{ji} & a_{jj} \end{bmatrix} =a_{ii}a_{jj}-a_{ji}a_{ij}=a_{ii}a_{jj}-a_{ij}^2<a_{kk}^2-a_{ij}^2 $$ What can we conclude?

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  • $\begingroup$ I don't understand why this determinant must be positive. The criterion (at least the one I know) says that the determinant of every $k \times k$ submatrix in the upper left corner of $A$ must be positive, but the matrix you consider is not a $2\times 2$ submatrix (unless $i=1$ and $j=2$). $\endgroup$ – user500094 Jan 17 '18 at 23:50
  • $\begingroup$ @user500094 If every leading principal minor is positive, then every principal minor is positive. $\endgroup$ – Brian Fitzpatrick Jan 18 '18 at 0:13

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