# Condition on Leading principle minors of a symmetric Positive semidefinite(PSD) matri

1. Apr 4, 2010

### NaturePaper

Hi everyone,
This is related to my previous https://www.physicsforums.com/showthread.php?t=392069"

Let $$A=(a_{ij})$$ be a symmetric (i.e., over reals) PSD matrix with the following conditions on Leading Principle Minors (determinant of the submatrix consisting of first i rows and i columns) $$A_{ii}$$:

$$A_{11}\ge0,~ A_{22}=A_{44}= A_{66}=A_{77}=A_{88}=detA=0$$

Now the question is can I say (from the above information) that $$A_{33}=A_{55}=0 ?$$ From "Matrix Analysis" by Horn and Johnson, I guess the Interlacing Inequlity may be useful...but I don't know much about it. Any help, please.

As usual, will it still valid if I assume A to be hermitian (i.e., over complex) than being symmetric?

Thanks

Last edited by a moderator: Apr 25, 2017
2. Apr 4, 2010

### NaturePaper

Re: Condition on Leading principle minors of a symmetric Positive semidefinite(PSD) m

Oh, I got the answer. It is indeed yes.

The result follows directly from Theorem 4.3.8 (page-185) of the book I mentioned above. It is a consequence of the "Interlacing inequality" as I guessed. Below is a brief sketch:

By our assumption, $$A_{22}$$ must have an eigenvalue 0 and hence by the interlacing property, the least eigenvalue (which can not be negative as it is PSD) of $$A_{33}\mbox{~is} \le0$$. Thus follows.

The result remains valid if A be hermitian.

More generally, we can say that for a hermitian PSD matrix $$A_{KK}=0~\Rightarrow~ A_{MM}=0~\forall M>K$$

Last edited: Apr 4, 2010