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I'm reading this paper and in the appendix I see the following statement:

For $A \in R^{m\times m}, B \in R^{n\times m}, C \in R^{n\times n}$,

if $D = \begin{bmatrix}A & B\\B^T & C\end{bmatrix}$ is positive semi-definite then,

$det(D) \leq det(A)det(C)$

This is given without proof as a property of psd matrices. This doesn't seem axiomatic to me and it's not obvious. Can you point to a reference or give a proof of this? I suspect it's pretty simple, but I'm missing it. I've never formally studied linear algebra so it might just be a gap in my education.

Some things I notice:

$A$ and $C$ are principal submatrices of $D$.

I know a determinant of a $2\times 2$ matrix is $a_{1,1}a_{2,2} - a_{1,2}a_{2,1}.$

Because $D$ is psd and has larger dimensions than $A$ or $C$, it seems like the second term is subtracting more than the second term for $A$ or $C$ would. But that statement is pretty imprecise and doesn't convince me that it's true.

munk
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1 Answers1

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Lemma: If $A$ and $B$ are symmetric positive-definite, then $\det(A+B) \geq \det(A)$.

This follows from Sylvester's determinant theorem: if $L$ is a Cholesky factor of $B$, $$\det(A+B) = \det(A)\det(I + L^TA^{-1}L) \geq \det(A)$$ since $L^TA^{-1}L$ is symmetric positive-definite, and adding $I$ shifts the spectrum by one.

Now for your original problem, if $D$ is singular the statement is obvious. If it's strictly positive-definite, write $$D=\left[\begin{array}{cc}A & 0\\B^T & I\end{array}\right]\left[\begin{array}{cc}I & A^{-1}B\\0 & C - B^TA^{-1}B\end{array}\right].$$ The matrix $C-B^T A^{-1}B$ is symmetric, and also positive-definite, since if $v$ is an eigenvector with negative eigenvalue $\lambda$, we have $$(-A^{-1}Bv\quad v)^T\,D\,(-A^{-1}Bv\quad v) = (-A^{-1}Bv\quad v)^T(0 \quad \lambda v) = \lambda\|v\|^2<0.$$ Thus by the lemma, $\det(C) \geq \det(C-B^TA^{-1}B)$, and your statement follows.

user7530
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