Sylvester's criteria and Negative definite Matrices. A)Sylvester's criterion states that a Hermitian matrix M is positive-definite if and only if all leading principal minors are positive.
AA) a Hermitian matrix M is negative-definite if and only if all leading principal minors are negative.
B)a Hermitian matrix M is positive-semidefinite if and only if all principal minors of M are nonnegative.
BB) a Hermitian matrix M is negative-semidefinite if and only if all principal minors of M are nonpositive.
Now My question is the following::
1) Can AA) be deduced from A) ? Or vice versa..
2) Can BB ) be deduced from B) ?  Or vice vera...
My Thoughts: I think they can be as if $A$ is positive definite or positive semi definite then $-A$ will be negative definite or negative semi definite.
So AA)[BB) ] can be deduced from  A)[BB) ].
Edit I wanted to ask if AA)[BB) ] can be deduced from A) [B) ]???? Basically I wanted to know if Sylvester's law is useful to determine whether a Matrix is negative-semidefinite or negative definite?? I am sorry..PLease edit your answer accordingly..
Can anyone please correct me if I went wrong anywhere??
Thank You.
 A: After long discussion I realised My thought was wrong. I was thinking that $det(-A) = - det(A)$. But this is possible only when $A$ is of odd order.
Now I will show how $A$ is negative definite if and only if all his LEADING principal minors of even order are positive and that of odd order are negative.
I assume that I know the Sylvester's criteria positive definite iff and only if all his LEADING principal minors are positive .
If $A$ is negative definite then $-A$ is positive definite.  All $(-A)$'s LEADING principal minors are positive Then all the even order leading principal minor of $A$ will give positive determinant while the odd ones will give negative . 
If all of $A$'s LEADING principal minors of even order are positive and that of odd order are negative. Then we can say all of $(-A)$'s Leading principal minor will be positive. So $-A$ is  positive definite. So $A$ is negative definite.
$A$ is negative semi definite iff and only if all his principal minors of even order are non-negative and that of odd order are non-positive.
proof will involve the same argument.
A: Let $A$ be a symmetric $ n \times n $ - matrix. For $k=0,1,,,n$ we denote the leading principal minors by $d_k$.
$A$ is positive definite $ \iff $ all $d_k>0$;
$A$ is negative definite $ \iff   (-1)^kd_k>0$ for $k=0,1,,,n$ .
A: I'll interpret your post as if there was an error in the line 

My Thoughts: I think they will be equivalent as if $A$ is positive definite or positive semi definite then $A$ won't be negative definite or negative semi definite.

Obviously these statement are mutually exclusive. A positive definite matrix is definitely not a negative definite, semidefinite or positive semidefinite matrix can only be one of them. Said so, you are saying that the statement are equivalent in the sense that: 

A matrix $A$ is positive definite $\Leftrightarrow$ $A$ is not negative definite

but this is most definitely not true because if a matrix is not, say, positive definite it can be either: negative definite, negative semidefinite or positive semidefinite! So that is why we have to have a definition for all four cases. Hope this clears your ideas 
