Let $A,B,C$ be n-order matrices. If $$M=\left(
\begin{matrix}
A &B
\\B^{T} &C
\end{matrix}
\right)$$ is a positive definite matrix, prove that $|M|\leq|A||C|$.
My attempt: Since A and C is necessarily also positive definite, there is some invertible $P$ and $Q$ such that $P^{T}AP=E_n$ and $Q^{T}CQ=E_n$, hence we have $$\left(
\begin{array}{cc}
P^{T} &0
\\0 &Q^{T}
\end{array}
\right)
\left(
\begin{array}{cc}
A &B
\\B^{T} &C
\end{array}
\right)
\left(
\begin{array}{cc}
P &0
\\0 &Q
\end{array}
\right)=\left(
\begin{array}{cc}
E_n &P^TBQ
\\Q^TB^TP &E_n
\end{array}
\right)$$ so taking det both sides we have $|M||A||C|\leq \det(RHS)$, but I'm stuck here. Does anyone know how to prove it? Thank you.
Best Answer
To finish the proof via congruence:
$P^TAP = I_n$ and $Q^TCQ = I_n$
$\implies \det\big(P^TP\big) = \det\big(A\big)^{-1}$and $\det\big(Q^TQ\big) = \det\big(C\big)^{-1}$
$\mathbf 0\prec \left[ \begin{array}{cc} P^{T} &\mathbf 0 \\\mathbf 0 &Q^{T} \end{array} \right] \left[ \begin{array}{cc} A &B \\B^{T} &C \end{array} \right] \left[ \begin{array}{cc} P &\mathbf 0 \\\mathbf 0 &Q \end{array} \right]=\left[ \begin{array}{cc} I_n &* \\* &I_n \end{array} \right]$
taking determinants:
$\det\left(\left[ \begin{array}{cc} P^{T} &\mathbf 0 \\\mathbf 0 &Q^{T} \end{array} \right]\right) \det\left(\left[ \begin{array}{cc} A &B \\B^{T} &C \end{array} \right]\right) \det\left(\left[ \begin{array}{cc} P &\mathbf 0 \\\mathbf 0 &Q \end{array} \right]\right)$
$=\det\left(\left[ \begin{array}{cc} P^{T}P &\mathbf 0 \\\mathbf 0 &Q^{T}Q \end{array} \right]\right) \cdot\det\big(M\big)$
$=\det\big(P^TP\big)\cdot \det\big(Q^TQ\big)\cdot\det\big(M\big)$
$=\det\big(A\big)^{-1}\cdot \det\big(C\big)^{-1}\cdot\det\big(M\big)$
$=\det\left(\left[ \begin{array}{cc} I_n &* \\* &I_n \end{array} \right)\right]$
$\leq 1$
by Hadamard's Determinant Inequality.
So $\det\big(A\big)^{-1}\cdot \det\big(C\big)^{-1}\cdot\det\big(M\big)\leq 1$ and re-scaling each side by $\det\big(A\big)\cdot \det\big(C\big)$ gives the result.
note:
There is no need for $A$ and $C$ to be the same size. This proof runs verbatim when $M\succ \mathbf 0$ and $A$ and $C$ are square. With this view: a special case of this inequality occurs when $C$ is $1\times 1$, which was this question:
Inequality for a determinant
Via induction, said special case proves Hadamard's Determinant Inequality which is at the core of this problem.