Best way to determine positivity of matrix I have a matrix $$A=\frac 12 \begin{bmatrix} 1+k+c &m&0&n+b \\
m &1+k-c&n-b & 0 \\
0 & n-b&1-k-c & -m \\
n+b & 0 & -m & 1-k+c\end{bmatrix}$$ where $0 \leq b \leq 1, k=\sqrt{1-b^2}$ and $$c=\frac {v-k}{w}$$ where $v \leq w(1-k)+k$ and $0 \leq w,v \leq 1$. Here $m$ and $n$ are free parameters. I'm trying to find a pair of real numbers $(m,n)$ which ensure that $A$ is positive semi-definite. For a fixed $b,k,c \in \mathbb R$, what is the best way to determine some $m,n$ which make $A$ positive semi-definite? I've attempted calculating the eigenvalues with Mathematica but they are far too complicated.
 A: Drop the positive factor $\frac12$. By permuting the rows and columns of $A$, we see that $A$ is similar to
$$
B=\begin{bmatrix}
1+k+c &0 &m &n+b\\
0 &1-k-c &n-b &-m\\
m &n-b &1+k-c &0\\
n+b &-m &0 &1-k+c
\end{bmatrix}.
$$
Obviously,


*

*when $1-k-c<0$, $B$ is not positive semidefinite;

*when $1-k-c=0$, $B$ is PSD if and only if $m=0,\ n=b$ and $(1+c)^2-k^2\ge(n+b)^2$.


Now suppose $1-k-c>0$. Then the leading principal $2\times2$ submatrix $M=\operatorname{diag}(1+k+c,\ 1-k-c)$ is positive definite and its Schur complement $S$ is given by
$$
\begin{bmatrix}
1+k-c &0\\
0 &1-k+c
\end{bmatrix}
-
\begin{bmatrix}
m &n-b\\
n+b &-m
\end{bmatrix}
\begin{bmatrix}
1+k+c &0\\
0 &1-k-c
\end{bmatrix}^{-1}
\begin{bmatrix}
m &n+b\\
n-b &-m
\end{bmatrix}.
$$
Since $B$ is congruent to $M\oplus S$, the matrices $B$ and $A$ are PSD if and only if $S$ is PSD, i.e. if and only if the two diagonal entries and determinant of $S$ are nonnegative.
A: By changing the order of the coordinates $\left({x}_{1} , {x}_{2} , {x}_{3} , {x}_{4}\right) \mathop{\longrightarrow}\limits \left({x}_{1} , {x}_{4} , {x}_{2} , {x}_{3}\right)$, we see that the problem is equivalent to
the positivity of the matrix
$${A'} = \left[\begin{array}{cccc}1+k+c&n+b&m&0\\
n+b&1-k+c&0&{-m}\\
m&0&1+k-c&n-b\\
0&{-m}&n-b&1-k-c
\end{array}\right] = \left[\begin{array}{cc}K&m D\\
m D&L
\end{array}\right]$$
As $K$ and $L$ are the matrices of the same quadratic form restricted to
two dimensional spaces, it follows that a necessary condition is that $K$ and
$L$ are positive semi-definite. Reciprocally, if $K$ and $L$ are
positive semi-definite, we can choose $m = 0$ and ${A'}$ is positive semi-definite (if we only want one possible value of $m$).
The non-negativity of the four diagonal elements is equivalent to $\boxed{\left|c\right|  \leqslant  1-k}$
(because $k \in  \left[0 , 1\right]$). We suppose that this condition is satisfied.
The non-negativity of $K$ and $L$
is then equivalent to their determinant being non negative, i.e.
$$\left\{\begin{array}{rcl}{\left(1+c\right)}^{2}-{k}^{2}& \geqslant &{\left(n+b\right)}^{2}\\
{\left(1-c\right)}^{2}-{k}^{2}& \geqslant &{\left(n-b\right)}^{2}
\end{array}\right. \quad  \Longleftrightarrow  \quad  \left\{\begin{array}{rcccl}{-b}-\sqrt{{\left(1+c\right)}^{2}-{k}^{2}}& \leqslant &n& \leqslant &{-b}+\sqrt{{\left(1+c\right)}^{2}-{k}^{2}}\\
b-\sqrt{{\left(1-c\right)}^{2}-{k}^{2}}& \leqslant &n& \leqslant &b+\sqrt{{\left(1-c\right)}^{2}-{k}^{2}}
\end{array}\right.$$
We see that ${A'}$ can be non-negative if and only if we can choose $n$ in the intersection of these
two intervals, that is to say if
$$\boxed{{-b}+\sqrt{{\left(1+c\right)}^{2}-{k}^{2}}  \geqslant  b-\sqrt{{\left(1-c\right)}^{2}-{k}^{2}}}$$
Then $n$ can be any value between
$$\max  \left({-b}-\sqrt{{\left(1+c\right)}^{2}-{k}^{2}} , b-\sqrt{{\left(1-c\right)}^{2}-{k}^{2}}\right)$$
and
$$\min  \left({-b}+\sqrt{{\left(1+c\right)}^{2}-{k}^{2}} , b+\sqrt{{\left(1-c\right)}^{2}-{k}^{2}}\right)$$
Edit: Possible values of m
Assuming $K$ and $L$ are positive semi-definite by the above conditions,
the set of $m$'s for which ${A'}$ is positive semi-definite can be studied like so: let
$${Q}_{m} \left(X , Y\right) = \left[\begin{array}{cc}X^\top&Y^\top
\end{array}\right] {A'} \left[\begin{array}{c}X\\
Y
\end{array}\right] = {X}^{\top } K X+2 m {Y}^{\top } D X+{Y}^{\top } L Y \qquad  X , Y \in  {\mathbb{R}}^{2}$$
As $m \mapsto  {Q}_{m} \left(X , Y\right)$ is an affine function, the set of all $m$'s such
that ${Q}_{m} \left(X , Y\right)  \geqslant  0$ is an interval of $\mathbb{R}$ containing $0$.
It follows that
$$I = \left\{m \in  \mathbb{R} \colon  \forall  X , Y \colon  {Q}_{m} \left(X , Y\right)  \geqslant  0\right\}$$
is also an interval containing $0$, because it is the intersection of such intervals, and by remarking that
${Q}_{{-m}} \left(X , Y\right) = {Q}_{m} \left({-X} , Y\right)$ we see that $I = \left[{-M} , M\right]$ is
symetrical around $0$.
A further reasoning shows that ${A'}$'s determinant must be $0$ when
$m = \pm  M$ because otherwise the eigenvalues would all be positive and a small
change in $m$ would not change the positive semi-definitness of ${A'}$.
I used sympy to compute (using ${k}^{2} = 1-{b}^{2}$)
$$\det  \left({A'}\right) = {m}^{4}+{m}^{2} \left(2 {c}^{2}+2 {n}^{2}-4\right)+\left({-4} {b}^{2} {n}^{2}+8 b c n+{c}^{4}-2 {c}^{2} {n}^{2}-4 {c}^{2}+{n}^{4}\right)$$
It follows that $M$ must be a positive root of this biquadratic function, which leaves only few choices.
