Conditions on coefficients of positive semidefinite matrix with certain symmetries I have a real matrix, with certain symmetries, defined as
$
A = \left( {\begin{array}{*{20}{c}}
1-x&a&b&c\\
a&x&d&b\\
b&d&x&a\\
c&b&a&1-x
\end{array}} \right),
$
with $x,a,b,c,d \in \mathbb{R},{\rm{~ }}0 \le x \le 1$.
I want to obtain conditions on coefficients $x,a,b,c,d$ for the matrix to be positive semidefinite.
For the particular case $x=0$, I obtain with the Mathematica Reduce command the following conditions for the eigenvalues to be nonnegative:
$a = b = 0,{\rm{~}} -1 \le c \le 1,{\rm{ ~}}d = 0$.
In[1]:= A = {{(1 - x), a, b, c}, {a, x, d, b}, {b, d, x, a}, {c, b, 
   a, (1 - x)}}

Out[1]= {{1 - x, a, b, c}, {a, x, d, b}, {b, d, x, a}, {c, b, a, 
  1 - x}}

In[2]:= FullSimplify[Eigenvalues[A /. x -> 0]]

Out[2]= {1/2 (1 - c - d - Sqrt[4 (a - b)^2 + (1 - c + d)^2]), 
 1/2 (1 - c - d + Sqrt[4 (a - b)^2 + (1 - c + d)^2]), 
 1/2 (1 + c - Sqrt[4 (a + b)^2 + (1 + c - d)^2] + d), 
 1/2 (1 + c + Sqrt[4 (a + b)^2 + (1 + c - d)^2] + d)}

In[3]:= Reduce[
 1/2 (1 - c - d - Sqrt[4 (a - b)^2 + (1 - c + d)^2]) >= 0 && 
  1/2 (1 - c - d + Sqrt[4 (a - b)^2 + (1 - c + d)^2]) >= 0 && 
  1/2 (1 + c - Sqrt[4 (a + b)^2 + (1 + c - d)^2] + d) >= 0 && 
  1/2 (1 + c + Sqrt[4 (a + b)^2 + (1 + c - d)^2] + d) >= 0]

Out[3]= b == 0 && a == 0 && d == 0 && -1 <= c <= 1

Similarly, for the particular case $x=1~$ I get
$~a = b = 0,{\rm{~}} -1 \le d \le 1,{\rm{ ~}}c = 0$.
However, for a general $~0\le x \le1$, Mathematica takes weeks without giving an answer. I suspect that for $~0< x <1~$ the condition $a = b = 0~$ must be satisfied.
Do you think there is a way to obtain some conditions for the general case  $~0\le x \le1$? The general eigenvalues of matrix $A$ have the following expressions:
$
\frac{1}{2} \left(-\sqrt{4 (a-b)^2+(-c+d-2 x+1)^2}-c-d+1\right),\frac{1}{2} \left(\sqrt{4 (a-b)^2+(-c+d-2 x+1)^2}-c-d+1\right),\frac{1}{2} \left(-\sqrt{4 (a+b)^2+(c-d-2 x+1)^2}+c+d+1\right),\frac{1}{2} \left(\sqrt{4 (a+b)^2+(c-d-2 x+1)^2}+c+d+1\right).
$
 A: Essentially the same approach:
the real symmetric matrix $A$ is positive if and only if the polynomial in $\lambda$
$$\det(\lambda I +A)$$ has all coefficients $\ge 0$. The nice thing in this case is that this polynomial of degree $4$ factors into two polynomials of degree $2$. The equivalent condition is that both these two polynomials have positive coefficients. We get the conditions
$$1-(c+d)\ge 0\\
(1-x-c)(x-d)\ge (a-b)^2\\
1+c+d \ge 0\\
(1-x+c)(x+d)\ge (a+b)^2$$
They appear weaker than the ones obtained by @Delta-u:, but in fact they are equivalent. 
Since the characteristic polynomial factors in this way, our matrix should be similar (by an orthogonal matrix) to the block matrix  
$$\left( \begin{matrix} 1-x -c & a-b& 0& 0\\
a-b& x-d & 0 & 0 \\
0 & 0 & 1- x + c & a+b \\
0& 0& a+b & x + d \end{matrix} \right)$$ 
I don't see right now a simple transformation.
A: You have both $1-(c+d)-\sqrt{4(a-b)^2+(1-2x+d-c)^2}$ and  $1-(c+d)+\sqrt{4(a-b)^2+(1-2x+d-c)^2}$ non negative iff:
$$1-(c+d)\geq 0$$
$$(1-(c+d))^2 \geq 4(a-b)^2 +(1-2x+d-c)^2$$
the second condition can be simplified as:
$$((1-x)-c)(x-d) \geq (a-b)^2$$
In particular $(1-x)-c$ and $x-d$ have the same sign.
As the sum $(1-x)-c+(x-d)=1-(c+d)$ both terms are positive.
Finally $1-(c+d) \pm \sqrt{4(a-b)^2+(1-2x+d-c)^2} \geq 0$ iff $c\leq 1-x$, $d<x$ and $((1-x)-c)(x-d) \geq (a-b)^2$.
The same computation on the two other eigenvalues give the final conditions:
$$-(1-x) \leq c \leq (1-x)$$
$$-x \leq d \leq x$$
$$((1-x)-c)(x-d) \geq (a-b)^2$$
$$((1-x)+c)(x+d) \geq (a+b)^2$$
If $x=1$ or $x=0$ these conditions are exactly the ones you described, if $0 \leq x \leq 1$ there is maybe a simpler way to express them but it is worth noticing that this is no longer necessary to have $a=b=0$.
