1
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If $A$ is real symmetric matrix then

a)does not contain $0$ eigenvalue

b)at least one eignvalue positive.

pick correct statement

1)option a is correct

2)option b is correct

3)both option a and b is correct

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6
  • $\begingroup$ What's your choice? $\endgroup$
    – user356774
    Dec 30, 2016 at 8:30
  • $\begingroup$ i think option 3)both a and b is correct $\endgroup$ Dec 30, 2016 at 8:31
  • $\begingroup$ So you think it doesn't have $0$ as an eigenvalue. In other words, you think that the matrix being real and symmetric automatically means the determinant is non-zero. This is easily shown false by, for instance, the symmetric $1\times 1$ matrix $[0]$. $\endgroup$
    – Arthur
    Dec 30, 2016 at 8:36
  • $\begingroup$ @Halima.Khatun : This should help : math.stackexchange.com/questions/1469778/… $\endgroup$
    – user356774
    Dec 30, 2016 at 8:38
  • $\begingroup$ What is your personal involvment in the subject, besides "I think that" without any other explanation ? $\endgroup$
    – Jean Marie
    Dec 30, 2016 at 13:21

1 Answer 1

1
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Consider $A_1=\begin{bmatrix}{1}&{0}\\{0}&{0}\end{bmatrix},\; A_2=\begin{bmatrix}{-1}&{0}\\{0}&{-1}\end{bmatrix}.$

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  • 1
    $\begingroup$ Why not just $-A_1$? $\endgroup$
    – Arthur
    Dec 30, 2016 at 9:14
  • $\begingroup$ Of course. That is another (and optimized) option. :) $\endgroup$ Dec 30, 2016 at 9:41

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