# Positive semi-definite in the limit

Consider the $n\times n$ matrix $F$ defined by the following expression $$F=A-\varepsilon B$$ where $A$ is a constant matrix such that $a_{ij}=a>0$ for all $i,j$ and where $B$ is a symmetric matrix such that $b_{ij}\geq 0$ for all $i,j$.

I would like to find sufficient conditions on $A$ and $B$ such that $F$ is positive semi-definite when the scalar $\varepsilon>0$ is small enough.

Of course, this works if $B$ is negative semi-definite. It also works for $n=2$ if the diagonal elements of $B$ are 0. In this case $$F=\left[\begin{array}{cc} a & a-\varepsilon b\\ a-\varepsilon b & a \end{array}\right]$$is diagonally dominant for $\varepsilon$ small enough. Sadly I'm unable to generalize this zero-diagonal case to $n>2$.

• A sufficient condition is that $B$ is strictly negative definite on the orthogonal complement of the vector $(1,1,...,1)$. Jun 12, 2017 at 14:49

Let $P$ denote an orthogonal projection onto the hyperplane $H:\sum x_i=0$. If $A-\varepsilon B$ is positive semi-definite, so is $P(A-\varepsilon B)P=-\varepsilon PBP$, thus we get a necessary condition: $PBP$ should be non-positive definite.

If $PBP$ is negative definite on $H$ (that is, the quadratic form $(Bx,x)$ is negative definite on $H$: $(Bx,x)\leqslant -c\|x\|^2$), this is a sufficient condition, that is, then $A-\varepsilon B$ is actually non-negative definite. Indeed, take a vector $z=x+y$, where $x\in H$, $y\perp H$. We get $$((A-\varepsilon B)z,z)=n\|y\|^2-\varepsilon (Bx,x)-2\varepsilon (Bx,y)-\varepsilon (By,y)\geqslant (n-1/2)\|y\|^2+c\varepsilon \|x\|^2-2\cdot \varepsilon \cdot \|B\|\cdot\|x\|\cdot\|y\|\geqslant 0$$ if $\varepsilon >0$ is small enough. If $PB$ has non-trivial kernel on $H$, the situation is more delicate. Namely, we may write any vector as $z=x_1+x_2+y$, where $y\perp H$ as before; $x_1,x_2\in H$, $Bx_2\perp H$ and $(Bx_1,x_1)\leqslant -c \|x_1\|^2$. We have $(Az,z)=n\|y\|^2$ as before and $(Bz,z)=2(Bx_1,y)+2(Bx_2,y)+(By,y)+(Bx_1,x_1)$. There are two subcases:

1) $Bx_2=0$ always, that is, the kernel of $PB$ on $H$ is contained in the kernel of $B$. Then the same argument works.

This is already necessary and sufficient condition. Indeed, if

2) $Bx_2\ne 0$ for some $x_2$. Then for any fixed $\varepsilon>0$ taking $x_1=0$ and $y$ of suitable sign and very small we get $((A-\varepsilon B)(x_2+y),(x_2+y))<0$.

• Thanks a lot for this answer. I'm confused about something: how do we get $$-\varepsilon(By,y)-2\varepsilon(Bx,y)\geqslant-1/2\|y\|^{2}-2\cdot\varepsilon\cdot\|B\|\cdot\|x\|\cdot\|y\|?$$ In particular, how does $1/2$ show up in there? Jun 13, 2017 at 13:00
• For small enough $\varepsilon$ it holds, yes? Jun 13, 2017 at 13:39
• It seems that a similar argument would show that if $B$ is negative definite on $H$ and $\varepsilon$ is small enough than $F$ is positive definite. Is this also true? Jul 10, 2018 at 17:06