Proof of Hadamard's Inequality for a Positive Semidefinite Matrix

Throughout, $\mathbb{R}^{n}$ is Euclidean space, $x\cdot y$ is the dot product, $Bx$ is the matrix-vector product, and $0_{\mathbb{R}^{n}}$ is the origin, all components of which are $0$.

Step 0: squares are nonnegative. Let $t\in\mathbb{R}$. By the totality of the order of the ordered field $\mathbb{R}$, either $0\le t$ or $t\le0$. If $0\le t$, then claim 5 of Elementary Arithmetic in an Ordered Field, applied to the inequality $0\le t$ and the nonnegative factor $t$, gives $0\,t\le t\,t$, and $0\,t=0$ by Zero Products and Elementary Identities in a Field. If $t\le0$, then adding $-t$ to both sides by claim 3 of Elementary Order Arithmetic in an Ordered Field gives $0\le-t$, so the previous case gives $0\le(-t)(-t)$, and $(-t)(-t)=t\,t$ by Zero Products and Elementary Identities in a Field. In both cases $0\le t\,t$.

Step 1: the diagonal entries are nonnegative. Fix $i\in[n]$ and let $e\in\mathbb{R}^{n}$ be the vector with $e_{i}=1$ and $e_{j}=0$ for $j\in[n]$ with $j\ne i$. For $j\in[n]$ the summands of $(Be)_{j}=\sum_{k=1}^{n}B_{jk}e_{k}$ vanish for $k\ne i$, so claim 7 of Properties of Finite Sums gives $(Be)_{j}=B_{ji}$. The same claim applied to $e\cdot(Be)=\sum_{j=1}^{n}e_{j}(Be)_{j}$ gives

$$e\cdot(Be)=(Be)_{i}=B_{ii}.$$

Since $B$ is positive semidefinite, $0\le e\cdot(Be)$, so $0\le B_{ii}$. As this holds for every $i\in[n]$, claim 5 of Properties of Finite Products gives $0\le\prod_{i=1}^{n}B_{ii}$.

Step 2: the positive definite case. Suppose $B$ is positive definite. By claim 4 of Properties of the Order on the Natural Numbers we have $1\le n$, so Cholesky Factorisation of a Symmetric Positive Definite Real Matrix applies and provides a lower triangular real $n\times n$ matrix $L$, in the sense of The Determinant of a Triangular Matrix is the Product of its Diagonal Entries, with $0<L_{ii}$ for every $i\in[n]$ and $B=L\,L^{\top}$, where $L^{\top}$ is the transpose and the product is the matrix product.

The matrix $L^{\top}$ is upper triangular with $(L^{\top})_{ii}=L_{ii}$, since $(L^{\top})_{il}=L_{li}$ and $L_{li}=0$ whenever $l<i$. Hence The Determinant is Multiplicative, claims 1 and 2 of The Determinant of a Triangular Matrix is the Product of its Diagonal Entries, and claim 2 of Properties of Finite Products give

$$\det B=\det L\ \det\bigl(L^{\top}\bigr)=\Bigl(\prod_{i=1}^{n}L_{ii}\Bigr)\Bigl(\prod_{i=1}^{n}L_{ii}\Bigr)=\prod_{i=1}^{n}L_{ii}L_{ii}.$$

On the other hand, by Product of Real Matrices and Transpose of a Real Matrix,

$$B_{ii}=\sum_{k=1}^{n}L_{ik}\bigl(L^{\top}\bigr)_{ki}=\sum_{k=1}^{n}L_{ik}L_{ik}\qquad(i\in[n]),$$

and every summand is nonnegative by Step 0, so claim 6 of Properties of Finite Sums gives $L_{ii}L_{ii}\le B_{ii}$ for every $i\in[n]$. Since also $0\le L_{ii}L_{ii}$ by Step 0, claim 5 of Properties of Finite Products gives

$$0\le\prod_{i=1}^{n}L_{ii}L_{ii}\le\prod_{i=1}^{n}B_{ii},$$

which is the assertion.

Step 3: the remaining case. Suppose $B$ is not positive definite. By Symmetric, Positive Semidefinite, and Positive Definite Real Matrices and the trichotomy of the order of $\mathbb{R}$ there is then an $x\in\mathbb{R}^{n}$ with $x\ne0_{\mathbb{R}^{n}}$ and $x\cdot(Bx)\le0$; since $B$ is positive semidefinite we also have $0\le x\cdot(Bx)$, so $x\cdot(Bx)=0$. Claim 3 of Cauchy-Schwarz Inequality for a Positive Semidefinite Quadratic Form on $\mathbb{R}^n$ gives $Bx=0_{\mathbb{R}^{n}}$, that is,

$$\sum_{i=1}^{n}B_{ji}x_{i}=0\qquad\text{for every }j\in[n].$$

Because $B$ is symmetric and multiplication in $\mathbb{R}$ is commutative, $B_{ji}x_{i}=x_{i}B_{ij}$, so $\sum_{i=1}^{n}x_{i}B_{ij}=0$ for every $j\in[n]$. Moreover $x\ne0_{\mathbb{R}^{n}}$ means that $x_{k}\ne0$ for at least one $k\in[n]$. Claim 6 of Row Properties of the Determinant therefore gives $\det B=0$, and Step 1 gives $0\le\prod_{i=1}^{n}B_{ii}$, so the asserted inequalities hold.