Principal minors of sum of a matrix and a diagonal matrix Let $A$ be a matrix such that all principal minors of $A$ are positive and $D$ be a diagonal matrix with non-negative diagonal entries. I need to prove that all the principal minors of $A+D$ are positive. 
Can anyone give any idea or hint to prove this?!
 A: We will make essential use of the matrix determinant lemma. The key result is the following proposition. I've included a full proof, so try not to read past the initial proposition if you want to try your hand at it yourself.
Proposition: Let $A$ be an $n\times n$ matrix such that all principal minors are positive. Then the matrix
$$A + \lambda\mathbf{e}_k\mathbf{e}^\mathrm{T}_k$$ 
has positive determinant for all $\lambda \ge 0$.
Here $\mathbf{e}_k\in \mathbb{R}^n$ is the $k$th standard basis vector of $\mathbb{R}^n$.
Proof: This is a direct application of the matrix determinant lemma, which says that we have
$$\det(A + \lambda\mathbf{e}_k\mathbf{e}^\mathrm{T}_k) = \det(A) + \lambda[\mathrm{Adj}(A)]_{kk},$$
where $[\mathrm{Adj}(A)]_{kk}$ denotes the $kk$th entry of the adjugate matrix of $A$. Explicitly, we know that 
$$[\mathrm{Adj}(A)]_{kk} = C_{kk},$$
where $C_{kk}$ is the $kk$th cofactor of $A$. But this is a principal minor of $A$, which by assumption was positive. Therefore it follows that we have
$$\det(A + \lambda\mathbf{e}_k\mathbf{e}^\mathrm{T}_k) = \det(A) + \lambda[\mathrm{Adj}(A)]_{kk} \ge \det(A) > 0,$$
where the $\geq$ sign becomes an equality if and only if $\lambda = 0$. $\square$
Corollary: Let $A$ be an $n\times n$ matrix such that all principal minors are positive. Then the matrix
$$B=A + \lambda\mathbf{e}_k\mathbf{e}^\mathrm{T}_k$$ 
also has all principal minors positive for $\lambda \ge 0$.
Proof: Consider a principal minor $[B]_I$, where $I$ is the index set of the rows/columns retained when forming the minor. If $k\notin I$, then the principal minor has not changed, i.e. we have $[B]_I = [A]_I > 0$. If $k\in I$, then apply the proposition to the underlying submatrix. $\square$
Finally, we can prove your result.
Proposition: Let $A$ be an $n\times n$ matrix with all positive principal minors. Let $D$ be a non-negative diagonal matrix with diagonal entries $d_k$. Then $A+D$ has all principal minors positive.
Proof: We can simply update $A$ one step at a time using the corollary. First, since $A$ has all principal minors positive, it follows that 
$$A_1 = A + d_1\mathbf{e}_1\mathbf{e}_1^\mathrm{T}$$
also has all principal minors positive. Now apply the corollary again to
$$A_2 = A_1 + d_2\mathbf{e}_2\mathbf{e}_2^\mathrm{T},$$
and inductively, to
$$A_k = A_{k-1} + d_k\mathbf{e}_k\mathbf{e}_k^\mathrm{T},$$
and conclude that each $A_k$ has all positive principal minors. The desired result follows by noting that $A+D = A_n$. $\square$
A: This can be proved by mathematical induction on the sizes $n$ of the matrices. The base case is trivial. Suppose the statement is true for matrices of sizes up to $n-1$. Now consider matrices $A$ and $D$ of size $n$. By induction hypothesis, all proper principal minors of $A+D$ are positive. It remains to show that $\det(A+D)>0$.
Define $f(t)=\det(A+tD)$ for every $t\geq0$. By Jacobi's formula,
$$
f'(t)=\operatorname{trace}\left(\operatorname{adj}(A+tD)D\right).
$$
Note that $\operatorname{adj}(A+tD)$ has a positive diagonal, because its diagonal elements are proper principal minors of $A+tD$, which are guaranteed to be positive by induction hypothesis. Therefore $f'(t)$ is always nonnegative. Yet, $f(0)=\det A$ is positive. So, $f$ must stay positive on $[0,\infty)$. In particular, $\det(A+D)=f(1)>0$.
