# If square matrix $\mathbf{A} \succ 0$, then how to prove Fischer's and Hadamard's inequalities?

If $\mathbf{A} = \begin{bmatrix} \mathbf{A}_{1} & x \\ x^* & a_{nn}\end{bmatrix} \succ 0$ is positive definite, in which $\mathbf{A} \in M_n$ and $\mathbf{A}_{1} \in M_{n-1}$, then

1. Is it possible to prove that $a_{nn} \det\left( \mathbf{A}_{1} \right) \geq \det\left( \mathbf{A}_{1} \right) \: \left(a_{nn} - x^*\mathbf{A}_{1}^{-1} x \right)$?
2. Can we deduce Hadamard's inequality from the above inequality somehow, i.e., $\det\left( \mathbf{A} \right) \leq \prod \limits_{i=1}^{n} a_{ii}$?

• I don't understand your proof, but the inequality in question 1 is known as Fischer's inequality. By applying it recursively to the leading principal submatrices of $A$, you can prove Hadamard's inequality. – user1551 Apr 28 '18 at 12:26
1. $\mathbf A$ is positive definite, therefore $\mathbf A_1$ is also positive definite (Sylvester's criterion) and in turn, $\mathbf A_1^{-1}$ is positive definite and $x^\ast\mathbf A_1^{-1}x\ge0$.
2. $a_{nn}-x^\ast\mathbf A_1^{-1}x$ is the Schur complement of $\mathbf A_1$ in $\mathbf A$. Therefore $(a_{nn}-x^\ast\mathbf A_1^{-1}x)\det(\mathbf A_1)=\det(\mathbf A)$ and the inequality in question 1 simply says that $a_{nn}\det(\mathbf A_1)\ge\det(\mathbf A)$. The result now follows from mathematical induction on the size of $\mathbf A$.