Conditions on coefficients of positive semidefinite matrix with certain symmetries

Question

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). $

Answer 1

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 $a=b=0$.

Answer 2

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.