4
$\begingroup$

I am to prove that the a symmetric matrix $A$ is positive definite iff the leading principal minors of $A$ are positive.

The forward implication is clear. Since the eigenvalues of a SPD matrix are positive and real, and $\det(A)$ is the product of eigenvalues, it must follow that these are positive too.

However, can someone please help me with the backwards implication? I just don't know how to handle it.

$\endgroup$
  • $\begingroup$ Your idea for the forward implication is a start but only establishes that the determinant of $A$ in its entirety is positive, not that all (leading) principal minors are positive. Incidentally, I tweaked your wording to make it more technically sound. Feel free to rollback or improve my edit if it does not meet with your approval. $\endgroup$ – hardmath Nov 10 '15 at 16:34
  • $\begingroup$ But any principle leading matrix is again SPD so any leading principle matrix of $A$ also has real, postive eigenvalues, so that $\det(A_k)>0$ right? $\endgroup$ – Wessel de Zeeuw Nov 10 '15 at 16:45
  • $\begingroup$ Yes, I'm now satisfied! $\endgroup$ – hardmath Nov 10 '15 at 16:45
  • $\begingroup$ Oke great! So only the backwards implication now;) $\endgroup$ – Wessel de Zeeuw Nov 10 '15 at 16:49
  • 1
    $\begingroup$ A real matrix is Hermitian if and only if it symmetric. $\endgroup$ – hardmath Nov 10 '15 at 17:02
2
$\begingroup$

You can prove it easily with the (already mentioned) Cauchy interlacing theorem, the fact that the determinant is equal to the product of eigenvalues and that the (real) symmetric matrix is positive definite if and only if its eigenvalues are positive.

The statement is obviously true for $1\times 1$ matrices. Assume that "all leading principal minors of $A$ are positive implies $A$ is positive definite" is true for $k\times k$ matrices, $k\leq n-1$, and consider $A$ to be $n\times n$ with all leading principal minors positive. By the induction assumption, we know that the leading principal $(n-1)\times(n-1)$ submatrix of $A$ is positive definite. By the interlacing property, all "larger" $n-1$ eigenvalues of $A$ are positive up to (possibly) the smallest one. However, the smallest eigenvalue cannot be nonpositive since otherwise the determinant of $A$ would not be positive.

$\endgroup$
  • $\begingroup$ may you give the arguments with our using Cauchy interlacing theorem??@algebraic $\endgroup$ – David Nov 30 '15 at 6:09

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.