Positive semi-definite in the limit

Consider the $n\times n$ matrix $F$ defined by the following expression $$F=A-\epsilon B$$ 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 conditions on $A$ and $B$ such that $F$ is positive semi-definite when the scalar $\epsilon>0$ is small enough.

Of course, this works if $B$ is negative semi-definite but I am looking for more general conditions, if possible.

Some thoughts:

It's easy to explain the case in which $B$ is symmetric positive-semidefinite.

Write $A = a \mathbf 1\mathbf 1^T$, where $\mathbf 1$ is the column-vector of $1$s. Note that $$x^T(A - \epsilon B)x = x^TAx - \epsilon (x^TBx)$$ $A - \epsilon B$ is positive semidefinite if the above is non-negative for all vectors $x$. Consequently, we must have $x^TAx = 0 \implies x^TBx = 0$, which in turn means that $Ax = 0 \implies Bx = 0$. In other words, the nullspace of $B$ must include the nullspace of $A$.

From this, we may conclude that $B$ must also be a multiple of $\mathbf 1\mathbf 1^T$ (i.e. it must also be a constant matrix).

More generally, we can write $B = B^+ - B^-$ where $B^\pm$ are positive semidefinite with $B^+B^- = B^-B^+ = 0$ (in particular: $B^\pm$ are simultaneously diagonalizable). We have $$x^T(A - \epsilon B)x = x^T(A + \epsilon B^-)x - \epsilon x^TB^+x\\ = [a(\mathbf 1^Tx)^2 + \epsilon x^TB^-x] - \epsilon x^TB^+ x$$ It is necessary (but not sufficient) that we have $(A + \epsilon B^-)x = 0 \implies B^+ x = 0$ for sufficiently small $\epsilon$.

Among the vectors $x$ such that $\mathbf 1^Tx = 0$, this condition becomes $B^- x = 0 \implies B^+ x = 0$. This tells us that if $B$ has an eigenvector parallel to $\mathbf 1$, we must have $B^- = 0$, which reduces to our earlier case.

Note that there are some examples in which $B$ is not positive semidefinite. In particular, we can take $$B = \pmatrix{0&1\\1&0}, \quad a = 1$$ In general: I suspect that if all diagonal entries of $B$ are zero, then $B$ will satisfy the required condition.

• Thanks, this is useful. I also feel that if the diagonal of $B$ is zero then there should be $\epsilon$ small enough so that $F$ is positive semi-definite. If $n=2$ then this works since for $\epsilon$ small enough $$F=\left[\begin{array}{cc} a & a-\epsilon b\\ a-\epsilon b & a \end{array}\right]$$ is diagonally dominant but I'm unable to generalize the argument to $n>2$. – user_lambda Jun 11 '17 at 19:27