Is there a $4$-by-$4$, rank $3$, positive semidefinite matrix with $a_{ii}=3$, $|a_{12}|\neq 1$, and principal minors having minimal eigenvalue $1$? Could anyone help me search for a positive semidefinite matrix $\left(a_{i,j}\right)_{4\times4}$ of rank 3 with $a_{i,i}=3$
and its all 3 by 3 principal minor matrices having minimal eigenvalue $\lambda_{\min}=1$, but $\left|a_{1,2}\right|\ne1$, or could anyone explain why such a matrix would not exist?
Thanks a lot.
 A: Suppose $A$ is a rank-$3$ positive semidefinite $4\times4$ matrix such that the minimum eigenvalues of its principal $3\times3$ submatrices are equal to all $1$. We will prove that
$$
A=D\pmatrix{3&1&1&1\\ 1&3&-1&-1\\ 1&-1&3&-1\\ 1&-1&-1&3}D^\ast
$$
for some unitary diagonal matrix $D$ (and hence all of its off-diagonal entries have unit moduli and the answer to your question is negative). Note that
\begin{aligned}
A&:=\pmatrix{3&a&b&c\\ \overline{a}&3&d&e\\ \overline{b}&\overline{d}&3&f\\ \overline{c}&\overline{e}&\overline{f}&3}\\
&=\pmatrix{1&\frac a2&\frac b2&0\\ \frac{\overline{a}}{2}&1&\frac d2&0\\ \frac{\overline{b}}{2}&\frac{\overline{d}}{2}&1&0\\ 0&0&0&0}
+\pmatrix{1&\frac a2&0&\frac c2\\ \frac{\overline{a}}{2}&1&0&\frac e2\\ 0&0&0&0\\ \frac{\overline{c}}{2}&\frac{\overline{e}}{2}&0&1}
+\pmatrix{1&0&\frac b2&\frac c2\\ 0&0&0&0\\ \frac{\overline{b}}{2}&0&1&\frac f2\\ \frac{\overline{c}}{2}&0&\frac{\overline{f}}{2}&1}
+\pmatrix{0&0&0&0\\ 0&1&\frac d2&\frac e2\\ 0&\frac{\overline{d}}{2}&1&\frac f2\\ 0&\frac{\overline{e}}{2}&\frac{\overline{f}}{2}&1}\\
&=:X+Y+Z+W.
\end{aligned}
By assumption, $\lambda_\min\left(A(1:3,1:3)\right)=1$. Since $X(1:3,1:3)=\frac12\left(A(1:3,1:3)-I_3\right)$, we have $\lambda_\min\left(X(1:3,1:3)\right)=0$ and hence $X$ is positive semidefinite. Similarly, $Y,Z$ and $W$ are PSD too.
As $A$ is singular, it has an eigenvector $v$ in its null space. Since $X,Y,Z,W$ are PSD and
$$
0=v^\ast Av=v^\ast Xv+v^\ast Yv+v^\ast Zv+v^\ast Wv,
$$
we must have $v^\ast Xv=v^\ast Yv=v^\ast Zv=v^\ast Wv=0$ and in turn $Xv=Yv=Zv=Wv=0$. By relabelling the rows and columns of $A$ and by scaling $v$ if necessary, we may assume that $v=(x,y,z,1)^\top$. The matrix equations $Xv=0,Yv=0,Zv=0$ and $Wv=0$ can then be rewritten as four systems of linear equations
$$
\begin{array}{ll}
\begin{cases}
2x+ay+bz=0,\\
\overline{a}x+2y+dz=0,\\
\overline{b}x+\overline{d}y+2z=0,
\end{cases}
&
\begin{cases}
2x+ay+c=0,\\
\overline{a}x+2y+e=0,\\
\overline{c}x+\overline{e}y+2=0,
\end{cases}
\\
\\
\begin{cases}
2x+bz+c=0,\\
\overline{b}x+2z+f=0,\\
\overline{c}x+\overline{f}z+2=0,
\end{cases}
&
\begin{cases}
2y+dz+e=0,\\
\overline{d}y+2z+f=0,\\
\overline{e}y+\overline{f}z+2=0.
\end{cases}
\end{array}
$$
Rearrange these equations into four groups, namely, those equations with the constant term $2$, those with the term $2x$, those with the term $2y$ and those with the term $2z$:
\begin{align}
&\overline{c}x+\overline{e}y+2=\overline{c}x+\overline{f}z+2=\overline{e}y+\overline{f}z+2=0\\
&2x+ay+bz=2x+ay+c=2x+bz+c=0\\
&\overline{a}x+2y+dz=\overline{a}x+2y+e=2y+dz+e=0\\
&\overline{b}x+\overline{d}y+2z=\overline{b}x+2z+f=\overline{d}y+2z+f=0.
\end{align}
From each group of equations in the above, we obtain
\begin{align}
&\overline{c}x=\overline{e}y=\overline{f}z=-1,\tag{1}\\
&ay=bz=c=-x,\tag{2}\\
&\overline{a}x=dz=e=-y,\tag{3}\\
&\overline{b}x=\overline{d}y=f=-z.\tag{4}
\end{align}
Substitute $c=-x$ (from $(2)$) into $\overline{c}x=-1$ (from $(1)$), we get $|c|=|x|=1$. Similarly, we also have $|e|=|y|=1$ and $|f|=|z|=1$. It follows from $(2),(3)$ and $(4)$ that $|a|,|b|$ and $|c|$ are also equal to $1$.
It follows that if we replace $A$ by $D^\ast AD$ for some appropriate unitary diagonal matrix $D$, we may assume that $a=b=c=1$. Then $(2)$ implies that $(x,y,z)=(-1,1,1)$, and $(3),(4)$ give $d=e=f=-1$. Now we are done.
