Can a matrix be translated by a diagonal matrix to obtain a singular matrix? Let $A$ and $D$ be $k \times k$ matrices with entries in the nonnegative real numbers $\mathbb{R}_+$, where $D$ is diagonal. 
Does there exist a scalar $\lambda \ge 0$ such that $\det(A - \lambda D) = 0$? 
 A: Let us assume $\det(D) > 0$, as otherwise the statement is not true in general.
Notice that for the symmetric root $R = D^{1/2}$ we have
$$ \det(A - \lambda D) = \det(R)\det(R^{-1} A R^{-1} - \lambda I_n) \det(R). $$
Now, the function $\det(R^{-1} A R^{-1} - \lambda I_n)$ has a non-negative zero if and only if $M = R^{-1} A R^{-1}$ has a non-negative eigenvalue.
If $A$ is symmetric:
The matrix $M$ has a non-negative eigenvalue if and only if $M$ is not negative definite. Since $R^{-1}$ is invertible, that is exactly the case if $A$ is not negative definite. Since $A$ is component wise non-negative, $A$ can't be negative definite, as the diagonal elements would negative otherwise. Thus, we obtain following result:

Let $A,D$ be square, component wise non-negative, matrices.
  Assume $A,D$ are symmetric and $D$ is positive definite.
  Then, there exists a $\lambda\ge 0$ with $\det(A - \lambda D)=0$.

If $A$ is not symmetric:
That case is little complicated.
In case $n$ is odd, if $\det(A) \ge 0$, then for $\lambda\to\infty$ we have 
$$ \det(A-\lambda D) = -\lambda^n \det(D - A/\lambda) \to -\infty. $$
In case $n$ is even, if $\det(A) \le 0$, then  $\lambda\to\infty$ we have 
$$ \det(A-\lambda D) = \lambda^n \det(D - A/\lambda) \to \infty. $$
So in these two cases, $\det(A-\lambda D)$ has a non-negative zero. 
A: Hint:
$\det(A-\lambda D)$ is a polynomial in $\lambda$ of degree $k$.
