Proof of Determinants of Positive Definite Matrices: Positivity, the Bound $\log\det A\le\mathrm{tr}\,A-d$, Bounds under Pinching, and the Expansion of $\det(I+tB)$

Each result cited below is universally quantified over the data in its own statement. Throughout, for $i\in[d]$, $e_{i}$ is the standard basis vector of $\mathbb{R}^{d}$ of Real Matrices, Symmetric Matrices and the Semidefinite Ordering: Standing Notation §basis, so that the $i$th coordinate of $z\in\mathbb{R}^{d}$ is $z\cdot e_{i}$ and $\lVert e_{i}\rVert=1$; determinants are given by the formula of Determinant of a Real Square Matrix, which is the reading used in Row Properties of the Determinant and in the results citing it.

Step 0 (auxiliary facts). (a) For every real $d\times d$ matrix $M$ and every $i\in[d]$, $e_{i}\cdot(Me_{i})=M_{ii}$. For $\alpha\in[d]$, the $\alpha$th coordinate of $Me_{i}$ is $\sum_{j=1}^{d}M_{\alpha j}(e_{i})_{j}=M_{\alpha i}$ by Matrix-Vector Product and claim 7 of Properties of Finite Sums, since $(e_{i})_{j}=0$ for $j\ne i$ and $(e_{i})_{i}=1$. By claim 1 of Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$, $e_{i}\cdot(Me_{i})=(Me_{i})\cdot e_{i}$, which is the $i$th coordinate $M_{ii}$ of $Me_{i}$.

(b) If $n$ is a natural number and $a:[n]\to\mathbb{R}$ has positive values $a_{k}$, then $\prod_{k=1}^{n}a_{k}$ is positive and $\log\prod_{k=1}^{n}a_{k}=\sum_{k=1}^{n}\log a_{k}$. We show both for the products and sums up to $j$, by induction on $j\in[n]$. For $j=1$ both sides reduce to $a_{1}$ and $\log a_{1}$ by claim 1 of Properties of Finite Products and claim 1 of Properties of Finite Sums. If the assertion holds for $j$ and $S(j)\in[n]$, then by those same claims $\prod_{k=1}^{S(j)}a_{k}=\bigl(\prod_{k=1}^{j}a_{k}\bigr)a_{S(j)}$, which is positive by claim 5 of Elementary Order Arithmetic in an Ordered Field, and by the rule $\log(st)=\log s+\log t$ of The Natural Logarithm its logarithm is $\sum_{k=1}^{j}\log a_{k}+\log a_{S(j)}=\sum_{k=1}^{S(j)}\log a_{k}$.

(c) If $n$ is a natural number and $c\in\mathbb{R}$, then $\sum_{k=1}^{n}c=n\,c$, with $n$ read in $\mathbb{R}$ as in The Real Numbers: Standing Notation and Background §numbers: by The Canonical Map from the Natural Numbers to a Field the sum of $n$ terms equal to $1$ is the real number $n$, and $\sum_{k=1}^{n}c\cdot1=c\sum_{k=1}^{n}1$ by claim 3 of Properties of Finite Sums.

(d) If $n$ is a natural number and $a:[n]\to\mathbb{R}$, then $\bigl|\prod_{k=1}^{n}a_{k}\bigr|=\prod_{k=1}^{n}|a_{k}|$, by induction on the number of factors, using the recursion of claim 1 of Properties of Finite Products and claim 4 of Properties of the Absolute Value in an Ordered Field.

Claim 1. Let $A\in\mathcal{S}(d)$ be positive definite. By Cholesky Factorisation of a Symmetric Positive Definite Real Matrix there is a lower triangular real $d\times d$ matrix $P$ with $0<P_{ii}$ for every $i\in[d]$ and $A=P\,P^{\top}$. By The Determinant is Multiplicative and claim 8 of Row Properties of the Determinant, $\det A=\det P\,\det(P^{\top})=\det P\,\det P$. By claim 1 of The Determinant of a Triangular Matrix is the Product of its Diagonal Entries, $\det P=\prod_{i=1}^{d}P_{ii}$, which is positive by Step 0(b). Hence $0<\det A$ by claim 5 of Elementary Order Arithmetic in an Ordered Field.

Claim 2. Let $A\in\mathcal{S}(d)$ be positive definite. For $i\in[d]$, $e_{i}$ is not the origin (its $i$th coordinate is $1$), so $0<e_{i}\cdot(Ae_{i})=A_{ii}$ by Symmetric, Positive Semidefinite, and Positive Definite Real Matrices and Step 0(a). Moreover $A$ is positive semidefinite: $x\cdot(Ax)$ is positive for $x$ nonzero, and for $x$ the origin $Ax$ is the origin by claim 1 of Linearity, Compatibility with the Matrix Product, and a Norm Bound for the Matrix-Vector Product, whence $x\cdot(Ax)=0$ by claim 5 of Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$ (the origin being $0z$ for any point $z$, by Scalar Multiple of a Point of $\mathbb{R}^n$ and claim 1 of Zero Products and Elementary Identities in a Field, and $0\,(x\cdot z)=0$ by the same claim). By Hadamard's Inequality for a Positive Semidefinite Matrix, $\det A\le\prod_{i=1}^{d}A_{ii}$; both sides are positive, by Claim 1 and Step 0(b). Since $\log$ is strictly increasing on $(0,\infty)$ by claim 2 of Real Powers Through the Exponential, and Elementary Asymptotic Tools: Monotonicity, Null Sequences of Negative Powers, Exponential Domination, Integer Rounding, and Square-Root and Exponential Inequalities, Step 0(b) gives

By the logarithm bound of the entropy-function lemma, $\log A_{ii}\le A_{ii}-1$ for every $i$. Hence, by claim 1 of Comparison and Absolute Value Bounds for Finite Sums of Real Numbers, claim 2 of Properties of Finite Sums, Step 0(c) and Trace of a Real Square Matrix,

Claim 3. Step 1 (positive definiteness). For $z\in\mathbb{R}^{d}$ and $a\in\mathbb{R}$, claims 1 and 2 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum give $(aI_{d})z=a(I_{d}z)=az$, so $z\cdot((aI_{d})z)=a(z\cdot z)=a\lVert z\rVert^{2}$ by claim 5 of Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$ and claim 1 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$. By the definition of the ordering (The Positive Semidefinite Ordering on Symmetric Matrices), the hypothesis therefore says

If $z$ is not the origin, then $\lVert z\rVert\ne0$ by claim 3 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, so $\lVert z\rVert$ is positive, $\varepsilon\lVert z\rVert^{2}$ is positive by claim 5 of Elementary Order Arithmetic in an Ordered Field, and $0<z\cdot(Az)$ by claim 2 there. As $A\in\mathcal{S}(d)$, $A$ is positive definite.

Step 2 (upper bound). Taking $z=e_{i}$ and using $\lVert e_{i}\rVert=1$ and Step 0(a) gives $A_{ii}\le L$ for every $i\in[d]$. By Trace of a Real Square Matrix, claim 1 of Comparison and Absolute Value Bounds for Finite Sums of Real Numbers and Step 0(c), $\mathrm{tr}\,A\le d\,L$, and Claim 2 gives $\log\det A\le\mathrm{tr}\,A-d\le d\,L-d$.

Step 3 (the inverse). By Invertibility of Symmetric Positive Definite Matrices, $A$ is invertible and $A'=A^{-1}$ is symmetric and positive definite, so $A'\in\mathcal{S}(d)$ and $AA'=I_{d}$. Fix $i\in[d]$ and put $w=A'e_{i}$. By claim 2 of Linearity, Compatibility with the Matrix Product, and a Norm Bound for the Matrix-Vector Product and claim 2 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum, $Aw=(AA')e_{i}=I_{d}e_{i}=e_{i}$. By Step 0(a) and claim 1 of Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$, $A'_{ii}=e_{i}\cdot w=w\cdot(Aw)$. Step 1 with $z=w$ gives $\varepsilon\lVert w\rVert^{2}\le A'_{ii}$, and claim 3 of Properties of the Absolute Value in an Ordered Field with Cauchy-Schwarz Inequality for the Euclidean Dot Product gives $A'_{ii}=e_{i}\cdot w\le|e_{i}\cdot w|\le\lVert e_{i}\rVert\,\lVert w\rVert=\lVert w\rVert$. If $\lVert w\rVert=0$, then $A'_{ii}\le0<\varepsilon^{-1}$. Otherwise $\lVert w\rVert$ is positive, and multiplying $\varepsilon\lVert w\rVert^{2}\le\lVert w\rVert$ first by $\lVert w\rVert^{-1}$ and then by $\varepsilon^{-1}$, both positive by claim 7 of Elementary Order Arithmetic in an Ordered Field, gives $\lVert w\rVert\le\varepsilon^{-1}$ (claim 5 of Elementary Arithmetic in an Ordered Field), hence $A'_{ii}\le\varepsilon^{-1}$. By Trace of a Real Square Matrix, claim 1 of Comparison and Absolute Value Bounds for Finite Sums of Real Numbers and Step 0(c), $\mathrm{tr}\,A'\le d\,\varepsilon^{-1}$.

Step 4 (lower bound). By The Determinant is Multiplicative and claim 1 of Row Properties of the Determinant, $\det A\,\det A'=\det(AA')=\det I_{d}=1$. Both determinants are positive by Claim 1, and $\log 1=0$ by claim 2 of Real Powers Through the Exponential, and Elementary Asymptotic Tools: Monotonicity, Null Sequences of Negative Powers, Exponential Domination, Integer Rounding, and Square-Root and Exponential Inequalities, so the rule $\log(st)=\log s+\log t$ of The Natural Logarithm gives $\log\det A=-\log\det A'$. Claim 2 applied to $A'$ and Step 3 give $\log\det A'\le\mathrm{tr}\,A'-d\le d\,\varepsilon^{-1}-d$. Hence, by sign reversal (claim 4 of Elementary Order Arithmetic in an Ordered Field), $d-d\,\varepsilon^{-1}\le-\log\det A'=\log\det A$.

Claim 4. Step 1 (the constants). Define real numbers $C'_{k}$, for natural numbers $k$, by recursion: $C'_{1}=0$ and $C'_{S(k)}=2C'_{k}+k$; by induction each $C'_{k}$ is nonnegative (claim 2 of Elementary Arithmetic in an Ordered Field, and $1\le k$ by claim 2 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field). Let $E_{d}=\prod_{l=1}^{d}2$ be the finite product of the constant family with value $2$, positive by Step 0(b). The set $S_{d}$ of permutations of $[d]$ is nonempty and finite by claim 4 of Finiteness of Cartesian Products, Tuple Sets, and Permutation Sets; let $N_{d}=\sum_{\sigma\in S_{d}}1$, the sum over the finite index set $S_{d}$, which is nonnegative by claim 2 of Nonnegativity and Monotonicity of a Sum over a Finite Index Set. Put

These depend only on $d$, and $C_{d}$ and $K_{d}$ are nonnegative. Since $1\le d$ (claim 2 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field) and $0\le C_{d}$, we have $1\le 2(d+C_{d})$; so $c_{d}$ is positive by claim 7 of Elementary Order Arithmetic in an Ordered Field, and multiplying $1\le2(d+C_{d})$ by $c_{d}$ (claim 5 of Elementary Arithmetic in an Ordered Field) gives $c_{d}\le c_{d}\cdot2(d+C_{d})=1$.

Step 2 (set-up). Let $B$, $m$ and $t$ be as in the statement with $|t|\,m\le1$, put $r=|t|\,m$, so $0\le r\le1$, and note $r^{2}=t^{2}m^{2}$ because $|t|^{2}=t^{2}$: by claim 1 of Properties of the Absolute Value in an Ordered Field, $|t|$ equals $t$ or $-t$, and $(-t)(-t)=t\,t$ by claim 2 of Zero Products and Elementary Identities in a Field. Multiplying $r\le1$ by $r$ gives $r^{2}\le r$. Let $M=I_{d}+tB$; by the entrywise definitions of the sum and the scalar multiple in Real Matrices, Symmetric Matrices and the Semidefinite Ordering: Standing Notation §matrices and Identity Matrix, $M_{ii}=1+tB_{ii}$ and $M_{ij}=tB_{ij}$ for $i\ne j$. By claim 4 of Properties of the Absolute Value in an Ordered Field and claim 5 of Elementary Arithmetic in an Ordered Field, $|tB_{ij}|=|t|\,|B_{ij}|\le r$ for all $i,j$. Hence, by claim 5 of Properties of the Absolute Value in an Ordered Field, $|M_{ii}|\le1+r\le2$, while $|M_{ij}|\le r\le1\le2$ for $i\ne j$.

Step 3 (the identity permutation). For $k\in[d]$ put $P_{k}=\prod_{i=1}^{k}M_{ii}$ and $Q_{k}=\sum_{i=1}^{k}tB_{ii}$. We show $|P_{k}-1-Q_{k}|\le C'_{k}r^{2}$ by induction on $k$. For $k=1$, $P_{1}-1-Q_{1}=(1+tB_{11})-1-tB_{11}=0$. Suppose the bound holds for $k$ and $S(k)\in[d]$, and put $x=tB_{S(k)S(k)}$. By claim 1 of Properties of Finite Products and claim 1 of Properties of Finite Sums, $P_{S(k)}=P_{k}(1+x)$ and $Q_{S(k)}=Q_{k}+x$, so

By claim 2 of Comparison and Absolute Value Bounds for Finite Sums of Real Numbers, claim 1 there, and Step 0(c), $|Q_{k}|\le\sum_{i=1}^{k}|tB_{ii}|\le k\,r$; so by claim 5 of Properties of the Absolute Value in an Ordered Field and $r^{2}\le r$, $|P_{k}-1|\le|P_{k}-1-Q_{k}|+|Q_{k}|\le C'_{k}r^{2}+k\,r\le(C'_{k}+k)\,r$. With $|x|\le r$, claim 4 of Properties of the Absolute Value in an Ordered Field and claim 5 of Elementary Arithmetic in an Ordered Field (applied twice) give $|(P_{k}-1)x|\le(C'_{k}+k)r^{2}$, and the triangle inequality gives $|P_{S(k)}-1-Q_{S(k)}|\le(2C'_{k}+k)r^{2}=C'_{S(k)}r^{2}$. In particular $|P_{d}-1-Q_{d}|\le C'_{d}r^{2}$, and $Q_{d}=t\sum_{i=1}^{d}B_{ii}=t\,\mathrm{tr}\,B$ by claim 3 of Properties of Finite Sums and Trace of a Real Square Matrix.

Step 4 (the other permutations). Let $\sigma\in S_{d}$ with $\sigma\ne\mathrm{id}$. There is $i\in[d]$ with $\sigma(i)\ne i$; put $j=\sigma(i)$, so $j\ne i$. Then $\sigma(j)\ne j$, for otherwise $\sigma(j)=j=\sigma(i)$ and injectivity of $\sigma$ would give $j=i$. For $l\in[d]$ let $\alpha'_{l}=r$ if $l=i$ and $\alpha'_{l}=1$ otherwise; $\alpha''_{l}=r$ if $l=j$ and $\alpha''_{l}=1$ otherwise; and $\beta_{l}=1$ if $l\in\{i,j\}$ and $\beta_{l}=2$ otherwise. By Step 2, $|M_{l\sigma(l)}|\le\alpha'_{l}\alpha''_{l}\beta_{l}$ for every $l$: for $l\in\{i,j\}$ we have $\sigma(l)\ne l$, so $|M_{l\sigma(l)}|\le r$, which equals the right side because $i\ne j$; for the other $l$ the right side is $2$. Hence, by Step 0(d), claim 5 of Properties of Finite Products, claim 2 there (twice) and claim 3 there,

the last step by claim 5 of Properties of Finite Products (as $0\le\beta_{l}\le2$) and claim 5 of Elementary Arithmetic in an Ordered Field. Since $\mathrm{sgn}(\sigma)$ is $1$ or $-1$ by claim 1 of The Sign of a Permutation is Multiplicative, also $\bigl|\mathrm{sgn}(\sigma)\prod_{l=1}^{d}M_{l\sigma(l)}\bigr|\le E_{d}r^{2}$.

Step 5 (the determinant). For $\sigma\in S_{d}$ let $T(\sigma)=\mathrm{sgn}(\sigma)\prod_{l=1}^{d}M_{l\sigma(l)}$, so that $\det M=\sum_{\sigma\in S_{d}}T(\sigma)$ by the formula recorded in Row Properties of the Determinant; by claim 1 of The Sign of a Permutation is Multiplicative, $T(\mathrm{id})=P_{d}$. Let $U(\mathrm{id})=1+t\,\mathrm{tr}\,B$ and $U(\sigma)=0$ for $\sigma\ne\mathrm{id}$; by claims 4 and 1 of Peeling, Splitting, and Interchange for Sums over a Finite Index Set, $\sum_{\sigma\in S_{d}}U(\sigma)=1+t\,\mathrm{tr}\,B$. Put $R(\sigma)=T(\sigma)-U(\sigma)$; by claims 3 and 4 of Properties of a Sum over a Finite Index Set,

By Steps 3 and 4, $|R(\mathrm{id})|\le C'_{d}r^{2}$ and $|R(\sigma)|=|T(\sigma)|\le E_{d}r^{2}$ for $\sigma\ne\mathrm{id}$; so $|R(\sigma)|\le(C'_{d}+E_{d})r^{2}$ for every $\sigma$. Let $n$ be the number of elements of $S_{d}$ and $\varphi:[n]\to S_{d}$ a bijection, as in Sum over a Finite Index Set; then claim 2 of Comparison and Absolute Value Bounds for Finite Sums of Real Numbers gives $\bigl|\sum_{\sigma\in S_{d}}R(\sigma)\bigr|=\bigl|\sum_{k=1}^{n}R(\varphi(k))\bigr|\le\sum_{k=1}^{n}|R(\varphi(k))|=\sum_{\sigma\in S_{d}}|R(\sigma)|$. By claim 1 of Nonnegativity and Monotonicity of a Sum over a Finite Index Set and claim 4 of Properties of a Sum over a Finite Index Set,

Step 6 (the logarithm). Suppose moreover $r\le c_{d}$, and put $s=\det M-1$. By claim 2 of Comparison and Absolute Value Bounds for Finite Sums of Real Numbers, claim 1 there, and claim 4 of Properties of the Absolute Value in an Ordered Field, $|t\,\mathrm{tr}\,B|\le|t|\sum_{i=1}^{d}|B_{ii}|\le|t|\,d\,m=d\,r$ (using Step 0(c)). By claim 5 of Properties of the Absolute Value in an Ordered Field, Step 5 and $r^{2}\le r$,

By claims 3 and 6 of Properties of the Absolute Value in an Ordered Field, $-\tfrac12\le s$, so $\det M=1+s\ge\tfrac12$ is positive; this is the first assertion. By the logarithm bound of the entropy-function lemma, applied with $1+s$ in place of $t$,

Here $1-(1+s)^{-1}=s-s^{2}(1+s)^{-1}$, and $(1+s)^{-1}\le2$. Indeed, since $-\tfrac12\le s$, multiplying by $2$, which is positive by claim 8 of Elementary Order Arithmetic in an Ordered Field, gives $-1\le 2s$ (claim 5 of Elementary Arithmetic in an Ordered Field), and adding $2$ gives $1\le2(1+s)$; the bound then follows by multiplying $1\le2(1+s)$ by the positive number $(1+s)^{-1}$ (claims 7 of Elementary Order Arithmetic in an Ordered Field and 5 of Elementary Arithmetic in an Ordered Field). Multiplying by $s^{2}\ge0$ gives $s^{2}(1+s)^{-1}\le2s^{2}$, so $-2s^{2}\le\log(1+s)-s\le0\le2s^{2}$, and $|\log\det M-s|\le2s^{2}$ by claim 6 of Properties of the Absolute Value in an Ordered Field. Since $s^{2}=|s|^{2}$ and $|s|\le(d+C_{d})r$, claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field gives $s^{2}\le(d+C_{d})^{2}r^{2}$. Finally, by claim 5 of Properties of the Absolute Value in an Ordered Field and Step 5,