# Positive semi-definiteness of a complex Matrix with special form.

Let $A,B,C$ complex $n\times n$ Matrices with ones on the diagonal and entries of absolute value 1. Further, let $$A=(a_{ij})_{i,j=1}^n,\; B=(b_{ij})_{i,j=1}^n \mbox{ and } C=(a_{ij}b_{ij})_{i,j=1}^n.$$

Can we say something about the positive semi-definiteness of $A$ if we know that $B$ and $C$ have this property?

Any hint to a similar problem or result or counter-example would be greatly appreciated.

• Are $A$, $B$, and $C$ related somehow? Why should positive semi-definiteness of $A$ depend at all on $B$ and $C$? – User8128 Jun 28 '17 at 17:43
• @User8128 Thank you. I forgot something ... edited the question – Vincent.W. Jun 28 '17 at 17:46
• Got it. Thanks makes more sense! – User8128 Jun 28 '17 at 17:46
• Not sure about the star notation. Is $C$ equal to the Hadamard product of $A$ and $B$? – Paul Aljabar Jun 28 '17 at 17:51
• @PaulAljabar Yes.. its just a componentwise product. – Vincent.W. Jun 28 '17 at 17:53

If by complex matrices you mean Hermitian matrices the answer is yes, for a silly reason and a non-silly reason. Suppose $B$ and $C$ are PSD with $1$s on the diagonal and all off diagonal entries of absolute value $1$. You ask if the elementwise quotient $A=(c_{i,j}/b_{i,j})$ is PSD. The silly reason is that the complex conjugate of a complex number of absolute value $1$ is its reciprocal. So the matrix $D=(\bar{b}_{i,j})$ is the matrix of reciprocals of the entries in $B$, and $A$ is the element-wise product of the two PSD matrices $A$ and $D$. Which it is also PSD, by the non-silly Schur product theorem: https://en.wikipedia.org/wiki/Schur_product_theorem.
• I mean, what is a complex positive definite matrix for you? Do you require $\bar z' M z\ge 0$ for all complex vectors $z$, with $\bar z'$ denoting the conjugate transpose? If so, the argument I supplied answers your question in the affirmative. – kimchi lover Jun 29 '17 at 15:02
• Yes, its just that you wrote that the answer is yes for hermitian matrices. But,if $B$ is psd and not hermitian then also its transpose conjugate $D$ will be psd. However then you dont have $B=D$ and hence $C\circ D\neq A$ ($\circ$ denotes hadamard) since $c_{i,j}*b^{-1}_{j,i}\neq a_{i,j}$. – Vincent.W. Jun 29 '17 at 15:44