0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Why does $A$ being symmetric positive definite imply that every leading principal minor of $A$ is positive? Beyond that , I completely understood, but I do not see how this comes about. $\endgroup$ – астон вілла олоф мэллбэрг Nov 9 '17 at 10:56
  • $\begingroup$ @астонвіллаолофмэллбэрг I edited, hope this is correct!:) $\endgroup$ – titusAdam Nov 9 '17 at 11:00
  • $\begingroup$ I am convinced. $\endgroup$ – астон вілла олоф мэллбэрг Nov 9 '17 at 11:01
  • $\begingroup$ Sylvester criterion seems to be a bit overkill here. One can prove it simpler by definition. Positive definite means $x^TAx>0$ for all nonzero $x$. Now take the truncated $x$ with the last $n-k$ coordinates being zero. $\endgroup$ – A.Γ. Dec 17 '18 at 13:25
-1
$\begingroup$

I think you are correct, but our professor prefers to use leading principal submatrix rather than leading principal minor, which sounds kind of broad.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.