Variables that will make this matrix positive semi-definite

Question

I have a matrix $$M=\begin{bmatrix} 1+t+m &n&t+n&m+c \\ n &1+t-m&m-c & t-n \\ t+n & m-c&1-t-m & -n \\ m+c & t-n & -n & 1-t+m \end{bmatrix}$$ where I know that $0 \leq c \leq 1$ and $ t=a-(m+n)b$ for some fixed $0 \leq a,b\leq 1$. Here $m$ and $n$ are free parameters with $t$ depending on $m,n$. I'm trying to find a pair of real numbers $(m,n)$ which ensure that $M$ is positive semi-definite. For a fixed $a,b,c \in \mathbb R$, what is the best way to determine some $m,n$ which make $M$ positive semi-definite? The eigenvalues of this matrix are $$\lambda=1 + (m+n) \pm \sqrt{c^2+m^2+n^2+2cm-2cn-2mn+2t^2}$$ and $$\lambda=1 - (m+n) \pm \sqrt{c^2+m^2+n^2-2cm+2cn-2mn+2t^2}.$$ If not, are there conditions on $a,b$ so that $m,n$ exist?

Answer 1

According to Laray's comment, a symmetric matrix is positive semi-definite iff all its eigenvalues are non-negative. The latter holds iff all the following conditions hold

(1) $|m+n|\le 1$

$(1 + (m+n))^2\ge (c^2+m^2+n^2+2cm-2cn-2mn+2t^2)$

$(1 - (m+n))^2\ge (c^2+m^2+n^2-2cm+2cn-2mn+2t^2)$.

After simplification, two latter inequalities transform to

(2) $1+4mn-c^2-2t^2\ge 2|cm-cn-m-n|$.

If we simultaneously change signs at $m$ an $n$ then all sides of inequalities 1 and 2 will remain the same, but when $m+n\ge 0$ then $|t|$ is not bigger than when $m+n<0$. So, without loss of generality we may assume that $m+n\ge 0$.

Substitute $u=m+n$ and $v=m-n$. Since $4mn=(m+n)^-(m-n)^2=u^2-v^2$ and $t=a-ub$, inequality 2 transforms to

(2’) $1+u^2-v^2-c^2-2(a-ub)^2\ge |cv-u|$.

Find $v_0$ for which $f(v)=v^2+|cv-u|$ attains its minimum. If $cv>u$ then $f(v)=v^2+cv-u$ increases when $v$ increases. So $cv_0\le u$. Then $f(v)=(v-\frac c2)^2-\frac {c^2}4+u$. Thus $v_0=\frac c2$ and $f(v_0)=u-\frac {c^2}4$ if $\frac {c^2}2\le u$; and $cv_0=u$ and $f(v_0)=v_0^2=\frac {u^2}c$, otherwise.

So the following cases are possible.

Case 1. $\frac{c^2}2\le u$. Then inequality 2’ transforms to $1+u^2-c^2-2(a-ub)^2\ge u-\frac{c^2}4$.

(2’.1) $g(u)=(1-2b^2)u^2+(4ab-1)u+1-2a^2-\frac {3}{4}c^2\ge 0$.

So the following cases are possible.

Case 1.1. $1-2b^2=0$, that is $b=\frac{\sqrt{2}}2$. When $4ab=2\sqrt{2}a\ne 1$ inequality 2’.1 becomes linear, so it holds for some $\frac{c^2}2\le u\le 1$ iff it holds for $u=\frac{c^2}2$ or $u=1$, that is when $4\sqrt{2}ac^2-7c^2-8a^2+4\ge 0$ or $2\sqrt{2}a-2a^2-\frac{3}{4}c^2\ge 0$. When $2\sqrt{2}a=1$ then $g(u)=1-2a^2-\frac{3}{4}c^2=\frac 34(1-c^2)\ge 0$ for any $u$.

Case 1.2. $1-2b^2\ne 0$, that is $b\ne\frac{\sqrt{2}}2$. Quadratic inequality 2’.1 holds for some $\frac{c^2}2\le u\le 1$ iff it holds for $u=\frac{c^2}2$ or $u=1$ or $u=\frac{1-4ab}{2-4b^2}$, provided this value belongs to the segment $\left[\frac{c^2}2,1\right]$.

After routine calculations, these conditions transform to that one of the following conditions is satisfied:

$(1-2b^2)c^4+(8ab-5)c^2+4-8a^2\ge 0$ (corresponds to $u=\frac{c^2}2$);

$1-2b^2+4ab-2a^2-\frac {3}{4}c^2\ge 0$ (corresponds to $u=1$);

$1-2a^2-\frac {3}{4}c^2-\frac{(1-4ab)^2}{4-8b^2}\ge 0$ and $\frac{c^2}2\le \frac{1-4ab}{2-4b^2}\le 1$ (corresponds to $u=\frac{1-4ab}{2-4b^2}$).

Case 2. $\frac {c^2}2> u$. Then inequality 2’ transforms to

$1+u^2-c^2-2(a-ub)^2\ge \frac {u^2}c$.

(2’.2) $g(u)=(c-2b^2c-1)u^2+4bcu-2a^2c-c^3\ge 0$

So the following cases are possible.

Case 2.1. $c-2b^2c-1=0$, that is $c=1$ an $b=0$. Then $g(u)=-2a^2-1<0$ for each $u$.

Case 2.2. $c-2b^2c-1<0$. Then $g(u)=-2a^2c-c^3<0$, and $g\left(\frac{c^2}2\right)=\frac{c}{4}( (-3-3b^2)c^2+8bc-8a^2)$. A polynomial $g(u)$ attains its maximum $\frac{4b^2c^2}{1+2b^2c-c}-2a^2c-c^3$ at a point $u_0=\frac{2bc}{1+2b^2c-c}\ge 0$. We have $u_0<\frac {c^2}2$ iff $-c^2+2b^2c -4b+c>0$. Thus in this case inequality 2’.2 holds for some $0\le u<\frac{c^2}2$ iff $u_0<\frac {c^2}2$ and $f(u_0)\ge 0$; or $u_0\ge\frac {c^2}2$ and $f(\frac {c^2}2)>0$. That is, iff $-c^2+2b^2c -4b+c>0$ and $\frac{4b^2c}{1+2b^2c-c}-2a^2-c^2\ge 0$; or $-c^2+2b^2c -4b+c\le 0$ and $ (-3-3b^2)c^2+8bc-8a^2\ge 0$.