Best way to determine positivity of matrix

Question

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.

Answer 1

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.

Answer 2

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.