Show that matrix $A_k$ is also symmetric and positive definite. Let $A\in\mathbb{R}^{n\times n}$ be a symmetric matrix that is positive definite.

Exercise: show that the matrix $A_k$ that consists of the first $k$ rows and columns of $A$ is also symmetric and positive definite:

My solution: We know that $A_k$ is symmetric because $A$ is symmetric. Furthermore, $A$ is SPD (symmetric positive definite) so every leading principal minor of $A$ is positive. This means that every leading principal minor of $A_k$ is positive, since $\{x:x \text{ is a leading principle minor of } A_k\} \subset \{y:y \text{ is leading principal minor of A}\}$. Because every leading principal minor of $A_k$ is positive, we have that $A_k$ is SPD.

EDIT: Matrix $A$ is SPD $\Rightarrow$ all its leading principal minors are positive: Let matrix $A$ be SPD and assume that there exists at least one leading principal minor that is nonpositive. This means that at least one of the eigenvalues of the corresponding leading principal submatrix is nonpositive and hence this leading principal submatrix is not SPD. This contradicts the fact that $A$ is SPD, since all the principal submatrices of a SPD matrix are SPD. So if $A$ is SPD all the leading principal minors are positive.

My question: is my solution correct or is it too intuitive? Should I be more elaborate when concluding that all leading principal minors of $A_k$ are positive?
I think you are correct, but our professor prefers to use leading principal submatrix rather than leading principal minor, which sounds kind of broad.
I think you are correct, but our professor prefers to use leading principal submatrix rather than leading principal minor, which sounds kind of broad.