If $A = [a_{ij}] \succeq 0 \in M_n$, and $a_{ii} a_{jj} = |a_{ij}|^2$, why $A$ is singular?

Question

If $A \in M_n$, $A \succeq 0$ is positive semidefinite, and $a_{ii} a_{jj} = |a_{ij}|^2$, then why $A$ is a non-invertible matrix?

Thank you in advance

p.s.: this problem is in "Matrix Analysis" by Horn and Johnson, second edition (please see [7.1.P1] page 434).

Answer 1

Not true. $$ \det\pmatrix{1&1&-1\\ 1&1&1\\ -1&1&1}=-4. $$

Edit. Under the new assumption that $A$ is positive semidefinite, the problem statement is true. Note that the $2\times2$ principal submatrix $A(\{i,j\},\,\{i,j\})$ is singular. Therefore, there exists some nonzero vector $v\in\mathbb C^2$ such that $v^\ast A(\{i,j\},\,\{i,j\})v=0$. In turn, by embedding $v$ in a longer zero vector, there exists a nonzero vector $x$ such that $x^\ast Ax=0$. Hence $A$ is singular.

Answer 2

The claim is true for $n=2$ and symmetric $A$, as then $\det A=0$ follows from the conditions on $a_{ij}$.

Answer 3

This is false for a general (possibly non-symmetric) positive semidefinite matrix: $$\begin{pmatrix}1 & -1 \\ 1 & 1\end{pmatrix}$$ is positive semidefinite (its associated quadratic form is $x^2+y^2$) yet invertible