1

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

Juniven Acapulco
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Halima.Khatun
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  • What's your choice? –  Dec 30 '16 at 08:30
  • i think option 3)both a and b is correct – Halima.Khatun Dec 30 '16 at 08:31
  • 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 $[0]$. – Arthur Dec 30 '16 at 08:36
  • @Halima.Khatun : This should help : http://math.stackexchange.com/questions/1469778/can-a-real-symmetric-matrix-have-0-zero-as-one-of-the-eigen-values –  Dec 30 '16 at 08:38
  • What is your personal involvment in the subject, besides "I think that" without any other explanation ? – Jean Marie Dec 30 '16 at 13:21
  • Neither option (a) nor (b) is necessarily true of all real symmetric matrices. First make sure you have understood definitions, then try simple examples to get a feeling for what is possible. When you understand the possibilities, then necessity will stand out more clearly for you. – hardmath Dec 30 '16 at 17:20

1 Answers1

1

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

Fernando Revilla
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