# 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.

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

• If A is positive-(semi)definite then -A is negative-(semi)definite. Is the statement true?@Fred – cmi Aug 14 '18 at 9:44
• Yes, it is true. – Fred Aug 14 '18 at 10:42
• Hii,, I have written an answer on my own . Can you please check my answer If I have gone wrong anywhere or not? @Fred – cmi Aug 14 '18 at 11:07

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.

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

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 $\det( \lambda A_{I,J})=\lambda^k\det A_{I,J}$. 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 $(-1)^{\lvert I\rvert}\det A_{I,I}\ge 0$ for all $I\in\mathcal P\{1,\cdots, n\}$.
• I can not understand your last line. I thought ------ If $A$ is non negative definite then it's every eigenvalue is non - negative. Then every eigen value of $-A$ is non positive. Am I not correct,@Saucy O'Path – cmi Aug 14 '18 at 9:30