A statement about positive semi-definite matrix

I heard that

for a Hermitian or symmetric matrix, if it is positive semi-definite, then all its square submatrices (not just those along diagonal) are non-negative

1. I suspect the statement is wrong.

There can be different interpretations of "non-negative".

If the statement meant to say "all its square submatrices (not just those along diagonal) have non-negative determinants", then matrix $[2,-1;-1,2]$ would be a positive semidefinite matrix but the square submatrix $[-1]$ doesn't have nonnegative determinant.

If the statement meant to say "all its square submatrices (not just those along diagonal) are positive semi-definite", then matrix $[2,-1;-1,2]$ would still be a positive semidefinite matrix but the square submatrix $[-1]$ isn't positive semi-definite.

So I wonder what the statement might actually want to mean? What is the closest correct statement to it?

2. Is the converse of the statement also true, i.e. "for a Hermitian or symmetric matrix, if all its square submatrices (not just those along diagonal) are non-negative, then it is positive semi-definite"?

Thanks!

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I think it wants to mean that for any set of indices $J\subset\{1,2,..,n\}$, the matrix $\ ( a_{jk})_{j,k\in J}\$ is positive semi-definite.
 Thanks! Is the converse true that "for a Hermitian or symmetric matrix, of $n$ by $n$, if for any set of indices $J⊂{1,2,..,n}$, the matrix $(a_{jk})j,k∈J$ is positive semi-definite, then the matrix is positive semi-definite"? – Tim Sep 30 '12 at 18:05 Added: in the condiition in my previous comment, I meant those submatrices that are not the full matrix. – Tim Sep 30 '12 at 19:04 Also is it true that "for a Hermitian or symmetric matrix, of n by n, if for any set of indices $J⊂1,2,..,n$, the matrix $(a_{jk})j,k∈J$ has nonnegative determinant, if and only if the matrix is positive semi-definite"? – Tim Sep 30 '12 at 19:17