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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.

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  • $\begingroup$ The $1 \times 1$ matrix 0 is real, symmetric and has zero as an eigenvalue. $\endgroup$
    – copper.hat
    Oct 8, 2015 at 3:51
  • $\begingroup$ 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. $\endgroup$
    – Capublanca
    Oct 8, 2015 at 3:51
  • $\begingroup$ Wow, i didn;t think of that. thanks Comments on the secon part @copper.hat $\endgroup$
    – bit_by_bit
    Oct 8, 2015 at 3:52
  • $\begingroup$ @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 $\endgroup$
    – bit_by_bit
    Oct 8, 2015 at 3:54
  • $\begingroup$ Try the identity and a corresponding Jordan block with eigenvalues 1. Try $I$ and $2I$. $\endgroup$
    – copper.hat
    Oct 8, 2015 at 3:54

1 Answer 1

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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.

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  • $\begingroup$ Thanks! So, What can be stated about, "If two matrices have the same eigen values, they have the same characteristic polynomial." @user254665 $\endgroup$
    – bit_by_bit
    Oct 8, 2015 at 10:30
  • $\begingroup$ Not in 3 or more dimensions. Every polyn. of deg n is the char. polyn. of an nxn matrix. Consider p=(x-1)(x-1)(x-2) and q=(x-1)(x-2)(x-2).The zeroes of the char. polyn. are the eigenvalues. $\endgroup$ Oct 8, 2015 at 15:40

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