Proof of Determinant Bound for a Matrix Squeezed between a Negative Multiple of the Identity and Zero

Throughout, $\mathbb{R}^{n}$ is Euclidean space, $z\cdot w$ is the dot product, $Mz$ is the matrix-vector product, and $0_{\mathbb{R}^{n}}$ is the origin. Put $B=(-1)A$, a real $n\times n$ matrix with $B_{ij}=-A_{ij}$; it is symmetric, since $B_{ij}=-A_{ij}=-A_{ji}=B_{ji}$. Note that $(-1)B=A$, because $(-1)(-A_{ij})=A_{ij}$.

Step 1: $B$ is positive semidefinite and $z\cdot(Bz)\le\lambda\,(z\cdot z)$. Let $z\in\mathbb{R}^{n}$. By claim 1 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum and Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$,

For every $j\in[n]$ the entries of $0_{n}$ vanish, so $(0_{n}z)_{j}=\sum_{k=1}^{n}0\,z_{k}$ has all summands equal to $0$ by Zero Products and Elementary Identities in a Field, and claim 7 of Properties of Finite Sums gives $(0_{n}z)_{j}=0$; the same two facts give $z\cdot(0_{n}z)=0$. Hence $A\preceq0_{n}$ means $z\cdot(Az)\le0$, so $0\le z\cdot(Bz)$ by claim 3 of Elementary Order Arithmetic in an Ordered Field. As $B$ is symmetric, $B$ is positive semidefinite in the sense of Symmetric, Positive Semidefinite, and Positive Definite Real Matrices.

By claims 1 and 2 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum we have $((-\lambda)I_{n})z=(-\lambda)z$, so Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$ gives $z\cdot(((-\lambda)I_{n})z)=(-\lambda)(z\cdot z)$. Hence $(-\lambda)I_{n}\preceq A$ means

and adding $\lambda\,(z\cdot z)+z\cdot(Bz)$ to both sides, by claim 3 of Elementary Order Arithmetic in an Ordered Field, gives $z\cdot(Bz)\le\lambda\,(z\cdot z)$.

Step 2: the diagonal entries of $B$ lie between $0$ and $\lambda$. Fix $i\in[n]$ and let $e\in\mathbb{R}^{n}$ be the vector with $e_{i}=1$ and $e_{j}=0$ for $j\ne i$. 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 $e\cdot(Be)=B_{ii}$, and applied to $e\cdot e=\sum_{j=1}^{n}e_{j}e_{j}$ gives $e\cdot e=1$. Step 1 with $z=e$ therefore yields

Step 3: bounding $\det B$. By Hadamard's Inequality for a Positive Semidefinite Matrix and by claim 5 of Properties of Finite Products, applied with the factors $B_{ii}$ and the constant factors $\lambda$, and by Natural Number Power of an Element of a Field,

Step 4: conclusion. Since $A=(-1)B$, claim 7 of Row Properties of the Determinant gives $\det A=(-1)^{n}\det B$. By claims 3 and 2 of Properties of Natural Number Powers in a Field and the identity $(-1)(-1)=1$ of Zero Products and Elementary Identities in a Field,

so, writing $x=(-1)^{n}$, the elementary field identities of Zero Products and Elementary Identities in a Field turn $xx=1$ into $(x-1)(x+1)=0$, and that lemma gives $x=1$ or $x=-1$.

By claim 5 of Properties of Natural Number Powers in a Field we have $0\le\lambda^{n}$, so claim 3 of Elementary Order Arithmetic in an Ordered Field gives $-\lambda^{n}\le0$. If $x=1$ then $\det A=\det B$ and Step 3 gives $-\lambda^{n}\le0\le\det A\le\lambda^{n}$. If $x=-1$ then $\det A=-\det B$, and negating the inequalities of Step 3 by claim 3 of Elementary Order Arithmetic in an Ordered Field gives $-\lambda^{n}\le\det A\le0\le\lambda^{n}$. In both cases

and claim 6 of Properties of the Absolute Value in an Ordered Field gives $|\det A|\le\lambda^{n}$.