A symmetric matrix A ∈ Mn(ℝ) is positive definite if and only if its leading principal minors are positive, that is, det Ai > 0 for i = 1,2,…,n.

See also,

 R.A. Horn, C.R. Johnson, Matrix Analysis, Cambridge University Press, Cambridge, 1996.

7.2.5 Theorem. If A ∈ Mn is Hermitian, then A is positive definite if and only if det Ai > 0 for i = 1,2,…,n.

As Wiwat explained, the leading principal minors should be positive, i.e. their determinants are positive

Some special cases are known. So if A_12 = A_21 is symetric, then A is positive semi definite if  A_12 is less than or equal to (PSD partial order) the geometric mean of A_11 and A_22. For definition of geometric mean see Ando. Its not just Sqrt[A__11 A_22] unless A_11 and A_22 commute

The schur complement theorem can solve your question. The block matrix A=[A11 A12;A21 A22] is symmetric positive definite matrix if and only if A11>0 and A11-A12^T A22^-1 A21>0.

I also agree with Cui

sorry i have a "same" question for this problem. Can anybody tell me why if i have symmetric matrix block A with rank(A) = rank(A11), then schur complement A11-A12^T A22^-1 A21 = 0?

Thanks in advance