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.
For a general $3\times 3$ matrix,
$$A =\begin{bmatrix}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}
\end{bmatrix}
$$ there is one third order principal minor namely $|A|$.
There are three second order principal minors:
$\begin{vmatrix}
a_{11} & a_{12}\\
a_{21} & a_{22}
\end{vmatrix}$ formed by deleting column $3$ and row $3$.
$\begin{vmatrix}
a_{11} & a_{13}\\
a_{31} & a_{33}
\end{vmatrix}$ formed by deleting column $2$ and row $2$.
$\begin{vmatrix}
a_{22} & a_{23}\\
a_{32} & a_{33}
\end{vmatrix}$ formed by deleting column $1$ and row $1$.
There are three first order principal minors: $|a_{11}|, |a_{22}|, a_{33}|$ .
Best Answer
This is a well known result and can be found, for example, here.
You argument is not quite correct, but can be saved if you use some elementary row transformations (which doesn't affect the rank) in order to make the other $n-k$ rows equal to zero. Then note that the corresponding column transformations also make the corresponding columns equal to zero. Now use your argument.