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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?

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  • $\begingroup$ Isn't this a reposting of your previous question? $\endgroup$ – user1551 Sep 18 '17 at 9:06
  • $\begingroup$ @user1551 There is an error in my previous question, the eigenvalues are incorrect. With the correct form of the eigenvalues the approach to answering this problem is quite different as it is not clear which one is the minimum. $\endgroup$ – gene Sep 18 '17 at 14:53
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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$.

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    $\begingroup$ Positive semi-definite is exactly, when all eigenvalues are non negative. That is the definition, I was told $\endgroup$ – Laray Sep 20 '17 at 7:44
  • $\begingroup$ @Laray Thanks.. $\endgroup$ – Alex Ravsky Sep 20 '17 at 8:30
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    $\begingroup$ Thanks, for completeness, if you finish the answer I will award the bounty $\endgroup$ – gene Sep 22 '17 at 14:38
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    $\begingroup$ @Alex If $a=b$ then we take $m=n=\frac{1-2ab}{2-4b^2}=\frac 12\ne\frac 14$, which results eigenvalues $-1,1,1,3$. Thanks, I’ll check my solution. $\endgroup$ – Alex Ravsky Sep 26 '17 at 6:06
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    $\begingroup$ @AlexRavsky There is something off about the assumption that you can take $m=n$. Take again, $a=b=1/2$ and $c=1$. None of your conditions are satisfied, however, taking $m=1$ and $n=0$ results in eigenvalues $4,0,0,0$. $\endgroup$ – Alex Sep 26 '17 at 17:11

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