Let $\mathbf{M}\in\mathbb{R}^{N \times N}$ be a symmetric positive definite matrix. Let us pick any integer $i$ and $j$ in $[1, N]$ with the only condition being that $i < j$. Finally, let us construct a $2\times 2$ "sub-matrix" named $\mathbf{A}_{ij}$ and defined as :
$$\mathbf{A}_{ij}=\begin{pmatrix}\mathbf{M}_{ii} & \mathbf{M}_{ij} \\ \mathbf{M}_{ij} & \mathbf{M}_{jj}\end{pmatrix}.$$
I have two questions :
- In this situation, is $\mathbf{A}_{ij}$ guaranteed to be positive definite or not ?
- Now a "reversed" situation : I don't know beforehand whether $\mathbf{M}$ is positive definite, but I know that every possible $\mathbf{A}_{ij}$ that one can construct are positive definite : does this imply then that $\mathbf{M}$ is positive definite ?
The matrix which you have constructed is a $2 \times 2$ principal submatrix. "Submatrix" is indeed the correct term, and this term is in common use. "Principal" here refers to the fact that the diagonals of your submatrix were taken from the diagonal of the original matrix (equivalently, the set of columns extracted and set of rows extracted are the same).
To your first question: yes. If $M$ is symmetric positive definite, then all of its $2 \times 2$ principal submatrices will be symmetric positive definite (in fact, the same can be said for all principal submatrices).
To your second question: no. One example of a matrix with positive definite $2 \times 2$ principal minors that fails to be positive definite is $$ M = \pmatrix{2&-1&3\\-1&5&0\\3&0&5} $$ In particular, we see that this $M$ fails to be invertible, but all of its $2 \times 2$ principal submatrices are invertible.
For a more dramatic example, the matrix $$ M = \pmatrix{10 & 3 & 2 & 1\\ 3 & 10 & 0 & 9\\ 2 & 0 & 10 & 4\\ 1 & 9 & 4 & 10} $$ is invertible, but has a negative determinant.