Suppose a $n \times n$ symmetric matrix $\mathbf{M}$ is positive definite.
Its block matrix form is written as follow.
\begin{align}
\mathbf{M} \; = \;
\begin{pmatrix}
\mathbf{A} & \mathbf{B}\\
\mathbf{B}^{T} & \mathbf{D}
\end{pmatrix}
\end{align}
Under what conditions can we say that the square matrices $\mathbf{A}$ or $\mathbf{D}$ are also positive definite?
Best Answer
As Algebraic Pavel stated correctly:
Here some elaborate statements:
If $A\in\mathbb{R}^{n\times n}$ is positive definite, then all of its principal submatrices $a_{1:m,1:m}$ ($m=1,\dots, n$) are positive definite. If $A$ is positive semi-definite, then all of its principal submatrices $a_{1:m,1:m}$ ($m=1,\dots, n$) are positive semi-definite.
This also works for negative (semi)-definite matrices, by simply multiplying the matrix by -1, i.e.
If $-A\in\mathbb{R}^{n\times n}$ is positive definite, then all of its principal submatrices $-a_{1:m,1:m}$ ($m=1,\dots, n$) are positive definite. If $-A$ is positive semidefinite, then all of its principal submatrices $-a_{1:m,1:m}$ ($m=1,\dots, n$) are positive semidefinite.
For a reference, see Observation 7.1.2 from Matrix Analysis (Horn, Johnson), 2nd edition.
See also Sylvester's criterion, which is a similar statement regarding principle minors.