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In this question matrix $A$ is positive-definite if and only if $\forall x\ne0 :x^TAx>0$. ($A$ is not necessarily symmetric)

Let $D$ be a positive-definite matrix such that it has block form:

$$D=\left( \begin{array}{cc} A & C \\ C^T & B \end{array} \right)$$

How can we prove that $\det D\leq\det A\det B$?

EDIT1: From the perspective of the first answer I want to sum up something.

  • It's true that

$$D=\left( \begin{array}{cc} A & C \\ C^T & B \end{array} \right)=\left( \begin{array}{cc} I & 0 \\ C^TA^{-1} & I \end{array} \right)\left( \begin{array}{cc} A & C \\ 0 & B-C^TA^{-1}C \end{array} \right)$$

so $\det D=\det A\det (B-C^TA^{-1}C)$.

It's also true that $\det A>0$ because every real eigenvalues of $A$ is greater than $0$ and all complex eigenvalues exist in pairs.

Now we need to prove that $$\det B>\det(B-C^TA^{-1}C)$$

  • The usual way when $A$ and $B$ are symmetric haven't worked yet because: (1) we don't know for sure if $(B-C^TA^{-1}C)$ is definite positive and (2) we don't know for sure if $\det(M+N)>\det(N)$ if $M$ and $N$ are definite positive.

I also want to point out that definite positive matrices (in this question) can have complex eigenvalues.

It would be great if you answer with details, not with references since almost every references consider positive-definite matrices to be symmetric.

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What you want to prove is not necessarily true. Here is a random counterexample:

$$ D=\left[\begin{array}{c|c}A&C\\ \hline C^T&B\end{array}\right] =\left[\begin{array}{rr|rr} 2&-3&1&0\\ 2&2&0&-1\\ \hline 1&0&2&1\\ 0&-1&-2&2 \end{array}\right]. $$ According to your definition, a matrix is "positive definite" if and only if its symmetric part is positive definite in the conventional sense. Now, WolframAlpha reckons that the eigenvalues of $$ D+D^T=\left[\begin{array}{rr|rr} 4&-1&2&0\\ -1&4&0&-2\\ \hline 2&0&4&-1\\ 0&-2&-1&4 \end{array}\right] $$ are $4-\sqrt{5}$ and $4+\sqrt{5}$ (each of multiplicity 2). Hence $D$ is "positive definite". WolframAlpha also reckons that $\det(D)=61$. However, $$\det(A)\det(B)=10\times6=60<\det(D).$$

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    $\begingroup$ congrat +500 such a releasing answer... ugmm I have been trying to disprove this so long $\endgroup$ – Leaning Sep 6 '14 at 14:59
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    $\begingroup$ @Mr.T Thanks. Apparently, the chance of getting a counterexample is small. I've just run a casual numerical experiment. If $C=I$ and the entries of $A$ and $B$ are taken from the range $[-10,10]$, then only 11% of those "positive definite" $D$s would violate the inequality in question. If the range is limited to $[-2,2]$, the proportion is even smaller -- 5.7%. So, if one tries only a few random examples without controlling the sizes of the entries of $A$ and $B$ carefully, one may not see any counterexample at all. $\endgroup$ – user1551 Sep 6 '14 at 15:59
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You need two facts to prove the result. One is about the determinant of the block matrices. $$\det D = \det A \det (B-C^\top A^{-1}C)$$ The other one is called Schur's complement. See page 66 here. $$D > 0 \leftrightarrow A > 0, B - C^\top A^{-1}C >0 $$

Note that $C^\top A^{-1}C \geq 0$ due to $A > 0$. This gives $B > 0$. From this you get $$\det (B-C^\top A^{-1}C) \leq \det B$$ Hence, $$\det D \leq \det A \det B$$

Edit: The determinant inequality follows from this

Edit #2 (Proof of $D > 0 \rightarrow A > 0 $): We assume $A$ to be square, $A \in \mathbb{R}^{n \times n}$. Note that when $A$ is square, it follows from the structure of $D$ that $B$ is square as well, $B \in \mathbb{R}^{m \times m}$. Take $x \in \mathbb{R}^{n}$. $$\begin{pmatrix} x^{\top} & 0 \end{pmatrix}D\begin{pmatrix} x \\ 0 \end{pmatrix} = x^{\top}Ax$$ Since $D$ is positive definite ($y^{\top}Dy >0 \quad \forall y \in \mathbb{R}^{(n+m) \times (n+m)}$), $x^{\top}Ax > 0 \quad \forall x \in \mathbb{R}^{n \times n}$. Hence, $A$ is positive definite. For the implication, $D > 0 \rightarrow B - C^\top A^{-1}C >0 $ I haven't been able to come up with a proof that doesn't require symmetricity assumption.

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  • $\begingroup$ can you explain a litter more explicitly? This is exactly the way I prove in the case all is symmetric. $\endgroup$ – Leaning Aug 31 '14 at 17:22
  • $\begingroup$ also in the mathoverflow site they assert the inequality for nonnegative HERMITIAN matrices only $\endgroup$ – Leaning Aug 31 '14 at 17:23
  • $\begingroup$ @Mr.T The Hermitian condition is for the Minkowski determinant theorem. There is another proof of the determinant inequality that doesn't make use of this assumption. It is in the comments section of the same answer there. $\endgroup$ – Calculon Aug 31 '14 at 17:28
  • $\begingroup$ @Mr.T None of the results here are limited to symmetric matrices. Both Schur's complement and the determinant identity hold under the assumptions you had in your question. $\endgroup$ – Calculon Aug 31 '14 at 17:30
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    $\begingroup$ @Mr.T You are right. I jumped too quickly to that conclusion. I'll try to fix it. $\endgroup$ – Calculon Sep 1 '14 at 7:48

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