# Can a real symmetric matrix have 0 (Zero) as one of the eigen values?

From what I know (correct me if I am wrong):

$$0$$ as an eigen value of a real symmetric matrix implies it is Singular (Non- invertible).

I am not aware of any such property with reference to real symmetric matrices.

Also, I wish to know if the following statements are correct or not.

a) If two matrices have the same eigenvalues, they have the same eigenvectors. (I think it's false)

b) If two matrices have the same eigen vectors, they have the same eigen values. (I think that's true)

Correct me.

• The $1 \times 1$ matrix 0 is real, symmetric and has zero as an eigenvalue. Oct 8, 2015 at 3:51
• The zero matrix (every entries is 0) is clearly symmetric, and it has $0$ as (the only) eigenvalue. The statements a) and b) are both false. Oct 8, 2015 at 3:51
• Wow, i didn;t think of that. thanks Comments on the secon part @copper.hat Oct 8, 2015 at 3:52
• @Capublanca can you elaborate on how both are false a bit? Does it not come from similarity? I am under the impression that if the eigen vectors/eigen values are same for 2 matrices, they are 'similar' too Oct 8, 2015 at 3:54
• Try the identity and a corresponding Jordan block with eigenvalues 1. Try $I$ and $2I$. Oct 8, 2015 at 3:54

Identify each of the following functions from $R^2$ to $R^2$ with their matrix representations : $f(x,y)=(x,0)$ , $g(x,y)=(2 x,0)$ , $h(x,y)=(0,y)$ , $i(x,y)=((x+y)/ 2,(x+y)/2)$. Now $f,g$ have the same eigenvectors but different eigenvalues. And $h,i$ have the same eigenvalues but different eigenvectors.