# If $A$ is real symmetric matrix [closed]

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

– user356774
Dec 30, 2016 at 8:30
• i think option 3)both a and b is correct Dec 30, 2016 at 8:31
• 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 $[0]$. Dec 30, 2016 at 8:36
• @Halima.Khatun : This should help : math.stackexchange.com/questions/1469778/…
– user356774
Dec 30, 2016 at 8:38
• What is your personal involvment in the subject, besides "I think that" without any other explanation ? Dec 30, 2016 at 13:21

Consider $A_1=\begin{bmatrix}{1}&{0}\\{0}&{0}\end{bmatrix},\; A_2=\begin{bmatrix}{-1}&{0}\\{0}&{-1}\end{bmatrix}.$
• Why not just $-A_1$? Dec 30, 2016 at 9:14