Edited per Ryan's clarification below.
Statement 1: Yes, this is fine. If $M$ is neither positive nor negative definite, and has no zero eigenvalues, then it must have at least one positive and one negative eigenvalue. Notice that this is a sufficient but not necessary condition on $M$ being indefinite. $\left[\begin{array}{ccc}0 & 0 &0\\0 & 1 & 0\\0 & 0 & -1\end{array}\right]$ is indefinite, for instance.
Statement 2: No, this is false. Consider for instance $\left[\begin{array}{cc}1 & 0\\0 & 0\end{array}\right]$ which is positive-semidefinite.
It is impossible to characterize indefinite matrices from the leading minors alone. For example, if the first row and column of a symmetric matrix $M$ is zero, the matrix might be positive-semidefinite, negative-semidefinite, or indefinite, yet all of the leading minors will be zero.
A complete, correct statement requires looking at all principal minors, for example: a symmetric matrix $M$ is indefinite (has positive and negative eigenvalues) if and only if:
- $\Delta_k < 0$ for some even $k$; or
- $\Delta_{k_1} > 0$ and $\Delta_{k_2} < 0$ for two different odd $k_1$ and $k_2$.
Knowing that $M$ is not strictly positive- or negative-definite does not really help. You can check that if $M$ satisfies neither of these conditions, then it must satisfy one of the rows of the purple box.
EDIT: Proof of the "only if" direction. Let $M$ be indefinite. Suppose, for contradiction, that neither of the above two hold. Then either all of the odd-dimensional minors are nonnegative, or all are nonpositive.
In the former case, $M$ satisfies the third row of the purple box above, and $M$ is positive-semidefinite, a contradiction.
In the latter case, $M$ satisfies the fourth row of the purple box above, and $M$ is negative-semidefinite, a contradiction.
EDIT 3: Proof of the "if" direction. Suppose one of the even-dimensional minors is negative, and suppose, for contradiction, that $M$ is positive-semidefinite or negative-semidefinite. Then by row three or four of the purple box (as appropriate), that minor is in fact positive, a contradiction. Therefore $M$ is neither positive- nor negative-semidefinite, and so is indefinite.
Suppose instead one of the odd-dimensional minors is positive, and another is negative, and suppose $M$ is positive-semidefinite. Then both of those minors are positive, a contradiction. Now suppose $M$ is negative-semidefinite. Then both of those minors are negative, a contradiction. The only remaining possibility is that $M$ is indefinite.
Best Answer
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 $A+D = A_n$. $\square$