Sign of eigenvalues of $A$ by $\det(A-\lambda I)=\lambda \det(B+D-\lambda I).$ Let $A$ be a $n\times n$ matrix, $B$ be a $(n-1)\times (n-1)$ matrix and $D$ be a $(n-1)\times (n-1)$ diagonal matrix with all entries positive. We assume that
$$\det(A-\lambda I)=\lambda \det(B+D-\lambda I)$$
and that all eigenvalues of $B$ are positive.
I was wondering if the eigenvalues of $A$ are all positive. I think yes, because $D$ are positive definite and
$$\det(A-\lambda I)=\lambda \det(B+D-\lambda I).$$
 A: This is not true. The matrix
$$
B=\pmatrix{ -2& 1\\ -22& 10}
$$
has real and positive eigenvalue (matlab says $0.2583$ and $7.7417$), but
$$
B+D=\pmatrix{ -1& 1\\ -22& 30}
$$
has a negative eigenvalue (eigenvalues are $-0.2733$ and $29.2733$).

The claim is true if in addition $B$ is symmetric, as Marcus M's answer shows.
A: Note that in the case of $\lambda = 0$, we have $\det(A - 0\cdot I) = 0$; this implies that $0$ is an eigenvalue of $A$, i.e. not all of the eigenvalues are positive.  
However, it is the case that all eigenvalues of $A$ are non-negative.  Since $B$ and $D$ are positive definite, then so is their sum, i.e. their sum only has positive eigenvalues.  Now, if $\lambda \neq 0$ is an eigenvalue of $A$, then $0 = \det(A - \lambda I) = \lambda \det(B + D -\lambda I)$.  Since $\lambda \neq 0$, this implies that $\det(B + D -\lambda I) = 0$, i.e. $\lambda$ is an eigenvalue of $B + D$.  However, since $B + D$ is positive definite, this implies that $\lambda > 0$.  Thus, all eigenvalues of $A$ are non-negative.
