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Symmetrix matrix $A\in\mathbb{R}^{n,n}$ is negatively defined. Then:
a. $\det_n(A)<0$
b. $a_{i,i} < 0$ for $i=1,2...,n$
c. matrix $A^2$ is positively defined.

a. it is not true, using fact about negative definition, for even $n=2k$ we have that $(-1)^{n}\det_{n}(A)>0\Leftrightarrow \det_n(A) >0$

b. I have no clue.

c. is not true,
$A= \left[ \begin{array}{ccc} -2 & 1 \\ 1 & -2 \end{array} \right]$
$\det_1(A) = -2$
$\det_2(A) = 3$

Am I ok ? What about corectness of a, c ? How to solve b ?
Edit - solution for b
We assume by contradiction: $a_{j,j} \ge 0$. Then I replace $j-th$ row with first row and $j-th$ column with first. Then $a_{1,1} \ge 0$. By these operations (replacing rows and column) don't change neither determinant nor minor, hence definition of change should be still negative. However, $\det A_{1} \ge 0$ so matrix can't be negatively defined. Contradiction.

2 Answers2

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Answer to question b is "yes" :

We rely on the fact that

$$A \ \text{negatively defined} \ \Longleftrightarrow (\forall X \in \mathbb{R}^n, X\neq 0 \Rightarrow X^TAX<0)$$

It suffice then to take $X^T=(0,0,\cdots 0,1,0, \cdots,0)$ (where the unique "$1$" is at index $i$) to have $X^TAX=a_{ii}<0.$

Jean Marie
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  • I edited my post, I added solution for b. Could you check my reasoning ? –  Aug 14 '16 at 21:10
  • What about c. ? –  Aug 14 '16 at 21:10
  • in fact c is true as @Mark L. Stone has explained. I don't understand your counterexample because if the original matrix is $A= \left[ \begin{array}{ccc} -2 & 1 \ 1 & -2 \end{array} \right]$ (which is neg. def.) then $A^2= \left[ \begin{array}{ccc} 5 & -4 \ -4 & 5 \end{array} \right]$ is pos. def. because $X^TAX=5x^2+5y^2-8xy=5(x-4y/5)^2+(9/5)y^2>0$ if $X^T= (x,y) \neq (0,0)$. – Jean Marie Aug 14 '16 at 21:28
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I will presume by "negatively defined" you mean negative definite. Note that a matrix A is negative definite if and only if -A is positive definite.

a. Looks o.k.

b. If a matrix is positive definite, what can you say about the sign of its diagonal elements? Well, then what can you say about the negative of the diagonal elements, which would be the diagonal elements of a negative definite matrix?

c. If a matrix is negative definite, all of its eigenvalues are negative. The eigenvalues of the square of a matrix are equal to the squares of the eigenvalues of the original matrix. Therefore, what can you conclude about the eigenvalues of the square of a negative definite matrix? Therefore what can you conclude as to whether or not the square of a negative definite matrix is positive definite?

Mark L. Stone
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  • If matrix is positevly defined then all eigenvalues are positive? –  Aug 14 '16 at 21:15
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    If a matrix is positive definite, then all its eigenvalues are positive. This is a very basic and important fact about positive definiteness. You should do some more reading of your textbook to better learn some of the basics. – Mark L. Stone Aug 14 '16 at 21:17
  • Ok, last question: Why The eigenvalues of the square of a matrix are equal to the squares of the eigenvalues of the original matrix. ? After all, it is not true that $\det(A^2 - I\lambda) = \det(A-I\lambda)$ –  Aug 14 '16 at 21:19
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    http://math.stackexchange.com/questions/227732/relationship-between-eigenvalues-of-a-matrix-and-its-square/227733 . There's no reason that your last equation should be true. – Mark L. Stone Aug 14 '16 at 21:21
  • Ok, so eigenvalues of $A^2$ are positive. But it doesn't say that $A^2$ is positively defined. –  Aug 14 '16 at 21:27
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    If eigenvalues of $A^2$ are positive, then $A^2$ is positive definite. – Mark L. Stone Aug 14 '16 at 21:45
  • So we have: eigenvalues are positvely defined if only and only matrix is positvely defined. ? –  Aug 15 '16 at 10:41
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    If we are talking about symmetric matrices, then yes. – Mark L. Stone Aug 15 '16 at 12:11
  • If matrix is not symmetric then only if $A$ is positevly defined then eigenvalues are also (strictly) positive –  Aug 15 '16 at 12:37
  • Something similar for negatively defined symmetric matrixes ? –  Aug 15 '16 at 12:43
  • Let us continue this discussion in chat. – Mark L. Stone Aug 15 '16 at 12:45