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$A=[a_{i,j}]\in\mathbb{R}^{n,n}$ is positive definit. Then:
a. $a_{i,j} > 0$ for $i,j=1,...,n$
b. $a_{i,i} > 0$ for $i=1,...,n$
c. matrix $A$ is nonsingular.

c. is true, it is easy thanks to Sylvester criterion. In particular, $\det A_n = \det A > 0$

a. is not true, counterexample: $ \left[ \begin{array}{ccc} 5 & -2 \\ -2 & 5 \\ \end{array} \right] $ is symettric and determinants = $5, 21$. So matrix is symmetric and positively definit, but $-2 < 0$.

b. it is false, conterexample:
$ \left[ \begin{array}{ccc} -5 & -2 \\ -2 & -5 \\ \end{array} \right] $

Am I ok?

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    $\begingroup$ You counterexample for b) is not valid, since your matrix is not positive definit. ($det A_1 = -5 < 0$) $\endgroup$ – Jakube Aug 5 '16 at 9:22
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It looks like pos def means strictly pos def.

For (a) ok. You have used Sylvester perfectly (alternatively you might notice that it is symmetric and calulate the evals [which are positive]).

For (b) it is not ok. Recall that (strictly) pos def means $x^T A x> 0$ for any non-zero vector $x$ and you may try with $x=e_i$ (canonical basis).

For (c) it is fine. A bit more details: pos def is quivalent to all the principal minors being positive. So in particular pos def implies that the full determinant is (strictly) positive so $A$ is non-singular. But your argument is fine.

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  • $\begingroup$ what about c. ? $\endgroup$ – user343207 Aug 5 '16 at 9:29
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    $\begingroup$ c is fine. By Sylvester pos def is equivalent to all principal minors having (strictly) pos dets. So in particular for the n'th as you state. $\endgroup$ – H. H. Rugh Aug 5 '16 at 9:32
  • $\begingroup$ When it comes to (a) I don't understand you. I shown symmetric matrix. Each principal determinants is strictly positive. Hence, matrix is positivly definit. As for (c) - could you edit your post and clear this issue ? $\endgroup$ – user343207 Aug 5 '16 at 17:33
  • $\begingroup$ Sorry, you are completely right about (a). Your response is perfect! I'll edit my post fro the rest. $\endgroup$ – H. H. Rugh Aug 5 '16 at 22:06

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