1
$\begingroup$

Let $A$ be a $n\times n$ real symmetric non-singular matrix. Suppose there exists $x\in \mathbb{R^n}$ such that $x'Ax<0$. Then we can conclude that

  1. $\det(A)<0$
  2. $B=-A$ is positive definite.
  3. $\exists y\in \mathbb{R^n}: y'A^{-1}y<0$
  4. $\forall y\in \mathbb{R^n}: y'A^{-1}y<0.$

My work:

I don't know how to do it for $n$, so I cooked up a $2\times 2$ matrix $\begin{bmatrix}-1 &2\\ 2& -1 \end{bmatrix}$, with $x=(1,0)^t$ we get $x'Ax<0$. Now $\det(A)=-3<0$ and also we can find a $y$ such that $y'A^{-1}y<0.$ But the problem is, according to the question only one option is true. And I don't know if this problem can be solved using a particular $n=2$.

So how can I solve this? Any help would be great. Thanks.

$\endgroup$
  • $\begingroup$ $1$ is not true, for example $-I$ wouldn't work. $2$ isn't true either, since a matrix could have mixed eigenvalue signs. Finally, the fourth one is the same as negative definiteness of $A^{-1}$, which again is a wrong statement. Hence it looks like the third one is correct. $\endgroup$ – астон вілла олоф мэллбэрг Oct 31 '16 at 12:35
  • $\begingroup$ What do you know about symmetric real matrices? $\endgroup$ – principal-ideal-domain Oct 31 '16 at 12:36
  • $\begingroup$ nothing much. Just definitions and some preliminary facts. $\endgroup$ – Kushal Bhuyan Oct 31 '16 at 13:11
1
$\begingroup$

Take $n=2$,

For Option-$1$: Consider $A=\begin{bmatrix}-1 & 0\\ 0& -1 \end{bmatrix}$ then clearly $A$ is symmetric and non-singular matrix.

Moreover, there exists $x=(1,0)$ such that $x^TAx=-1$ but $det A=1$.

Option-$4$: Consider same matrix then $A^{-1}=\begin{bmatrix}-1 & 0\\ 0& -1 \end{bmatrix}$ but for $y=(0,0), y^TAy=0$

Option-$2$: Consider $A=\begin{bmatrix}-1 & 0\\ 0& 1 \end{bmatrix}$ then there exists $x=(-1,0)$ such that $x^TAx<0$ but $B=-A=\begin{bmatrix}1 & 0\\ 0& -1 \end{bmatrix}$ is not positive definite as determinant of $B$ is $-1$.

Hence, Option-$3$ is true.

$\endgroup$
  • $\begingroup$ Elimination of other options is not good Idea to say remaining option is correct. Can you please give me hint that how can we prove third option is correct? $\endgroup$ – ramanujan Nov 30 '18 at 0:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.