# 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.

• This doesn't look fun. You can use Sylvester's criterion (en.wikipedia.org/wiki/Sylvester%27s_criterion) to write down inequalities and see if they are satisfied. If you are ready to settle down for positive definite matrix, you'll get four polynomial inequalities from the four leading minors. If you really want only positive semi-definiteness, you'll need to consider all the principal minors... – levap Sep 6 '17 at 6:29
• You can completely remove $v$ and $w$ from the problem's description because if the matrix is positive semi-definite then the third diagonal term implies $k\le1-c$ and this implies that there exists $v, w\in [0,1]^2$ such that $k = v - c w$, because $0 - c \cdot 0 \le k \le 1 - c\cdot 1$. Furthermore $v = c w + k \le (1-k) w + k$ is satisfied. – Gribouillis Sep 6 '17 at 7:22

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.

• Great answer, thank you. – gene Sep 6 '17 at 18:37
• What about when ${-b}+\sqrt{{\left(1+c\right)}^{2}-{k}^{2}} < b-\sqrt{{\left(1-c\right)}^{2}-{k}^{2}}$? – Alex Sep 7 '17 at 21:42
• @Alex Then whatever the value of $n$, one of the matrices $K$ or $L$ is not positive semi-definite. Hence $A^\prime$ cannot be positive semi-definite. – Gribouillis Sep 7 '17 at 21:45
• @Gribouillis But what if you take $m \neq 0$? Why do you assume that $m=0$? – Alex Sep 7 '17 at 22:19
• @Alex Whatever the value of $m$, $A^\prime$ cannot be positive semi-definite if $K$ and $L$ are not positive semi-definite. On the other hand when they are, the value $m=0$ always works. If $K$ and $L$ are positive definite, then at least a neighbourhood of $m = 0$ works, but my post does not study the set of possible values of $m$. – Gribouillis Sep 7 '17 at 22:23

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.