Vtyjkyuk

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.

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 $k=0,1,,,n$ .

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

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.

Careful there, because minors are not linear functions: a minor of order $k$ is a $k$-homogeneous function, id est: if $I=i_1,cdots, i_k$ and $J=j_1,cdots, j_k$ are subsets of $1,cdots,n$ with cardinality $k$, then $det( lambda A_I,J)=lambda^kdet A_I,J$. And you can see what happens to you criterion (B) when $lambda=-1$:

$A$ is negative-semidefinite if and only if $-A$ is positive-semidefinite, if and only if $(-1)^lvert Irvertdet A_I,Ige 0$ for all $Iinmathcal P1,cdots, n$.

Same with (A).