How many positive semidefinite submatrices can an indefinite matrix with a positive eigenvector/positive eigenvalue have? Let $A\in\mathbb R^{n\times n}$ be a symmetric matrix such that it has $k<n$ non-negative eigenvalues and $n-k$ negative eigenvalues. I am interested in the following question: can all $k\times k$ principal submatrices of $A$ be positive semidefinite? If not, how many of them can be positive semidefinite?
I was experimenting with Matlab for the case $n=4$, $k=2$, and I couldn't find a way to make $A$ so that all $2\times 2$ matrices were positive semidefinite (somehow $A$ always would get more than two non-negative eigenvalues). Hence the question. I can't figure it out myself.
Edit: there are two cases I already understand: $k=1$ and $k=n-1$.
In the first case it is easy to find a matrix with one positive eigenvalue such that the diagonal is non-negative: say, $A=\begin{bmatrix}0&1&1\\1&0&1\\1&1&0\end{bmatrix}$ (the ones on the off-diagonals may be any positive numbers). For the second case, consider $-A^{-1} = \begin{bmatrix}1&-1&-1\\-1&1&-1\\-1&-1&1\end{bmatrix}/2$. But what if $n>3,1<k<n-1$?
Another edit: one user suggested to use a rectangular $(n\times m)$ (with $m<n$) matrix $X$ such that every set of $k$ rows of $X$ has full rank (like a Vandermonde matrix) and look at $A=cXX^* - I_n$ for some $c>0$ chosen such that $A$ has $k$ non-negative eigenvalues and $k\times k$ PD submatrices. Indeed, this method was successful. So the question is answered, technically, but there is more: the matrix always seems to have a negative eigenvalue with a positive vector.
This leads to the same question, modified by the assumption: suppose that the subspace $V$ spanned by the eigenvectors corresponding to positive eigenvalues contains a positive (entrywise $>0$) vector.
 A: Let $X\in M_{n,k}(\mathbb R)$ be a rectangular Vandemonde
matrix such that $X_{ij}=x_i^{j-1}$ for some $n$ distinct positive real numbers $x_1,x_2,\ldots,x_n$. By construction, each $k\times k$ submatrix $Y$ of $X$ is nonsingular and entrywise positive.
Let $A=cXX^T-I_n$ for some sufficiently large $c>0$. Then $A$ has $k$ positive eigenvalues (namely, $c\sigma_i(X)^2-1$ for $i=1,2,\ldots,k$) and an eigenvalue $-1$ of multiplicity $n-k$. Moreover, all $k\times k$ principal submatrices of $A$ are in the form of $cYY^T-I_k$. Since $c\sigma_\min(Y)^2-1>0$ when $c$ is large, these principal submatrices are positive definite.
Also, when $c$ is large, $A$ and its submatrices are entrywise positive. Therefore, by Perron-Frobenius theorem, each of its principal submatrices (including $A$ itself) contains a positive eigenvector corresponding to a positive eigenvalue (the spectral radius of the submatrix).
A: Let $A$ have eigenvalues $\lambda_n \leq \cdots \leq \lambda_1$. By the min-max theorem, we have that
$$\lambda_k = \max_{\dim U = k} \min_{v \in U} \frac{v \cdot Av}{v \cdot v}.$$
For $I \subseteq \{1, \ldots, n\}$, let $\langle I \rangle \subseteq \mathbb{R}^n$ by the span of $\{e_i \mid i \in I\}$. Then for any $v \in \langle I \rangle$, we have that $v \cdot A v = v \cdot A_I v$, where $A_I$ is the principal $I$-submatrix of $A$. When $|I| = k$, then $\dim \langle I \rangle = k$ and hence by the min-max theorem we have
$$ \lambda_k \geq \min_{v \in \langle I \rangle} \frac{v \cdot A_I v}{v \cdot v} = \lambda^I_k, $$
where $\lambda^I_k$ denotes the smallest eigenvalue of $A_I$. Therefore, the smallest eigenvalues of any $k \times k$ principal submatrix are bounded above by $\lambda_k$.
By your assumption on $A$, $\lambda_k \geq 0$, so this doesn't say much about the definiteness of the $k \times k$ principal submatrices. However, you also have that $\lambda_{k + 1} < 0$, so every $(k + 1) \times (k + 1)$ principal submatrix is certainly not positive semidefinite. Perhaps you can do something more precise with the Cauchy interlacing theorem?
