Can a matrix be translated by a diagonal matrix to obtain a singular matrix?

Question

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 $\det(A - \lambda D) = 0$?

Answer 1

Hint:

$\det(A-\lambda D)$ is a polynomial in $\lambda$ of degree $k$.

Answer 2

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.