# A is a positive definite matrix iff its leading principal minors are positive

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.

• 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. – hardmath Nov 10 '15 at 16:34
• 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? – Wessel de Zeeuw Nov 10 '15 at 16:45
• Yes, I'm now satisfied! – hardmath Nov 10 '15 at 16:45
• Oke great! So only the backwards implication now;) – Wessel de Zeeuw Nov 10 '15 at 16:49
• A real matrix is Hermitian if and only if it symmetric. – hardmath Nov 10 '15 at 17:02

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.