LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A nonsingular principal minor of the symmetrized Cartan matrix of size the rank

Statement

Let $n \geq 1$ and let $B = (b_{i j})$ be a real symmetric $n \times n$ matrix of rank $r$ . For $J \subseteq {1, \dots, n}$ , write $B_{J} = (b_{i j})_{i, j \in J}$ for its principal submatrix. Use the convention that the empty matrix has determinant $1$ and is invertible. There is a subset $J$ with $∣ J ∣ = r$ for which $B_{J}$ is nonsingular.

If in addition $A$ is a real $n \times n$ matrix and $D = diag (d_{1}, \dots, d_{n})$ has $d_{i} > 0$ with $B = D A$ , then for the same $J$ ,

$\det B_{J} = (\prod_{i \in J} d_{i}) \det A_{J} .$

Thus $A_{J}$ is also nonsingular and has size $r$ , and $∣ {1, \dots, n} ∖ J ∣ = n - r$ . In particular, if $A$ is symmetrizable and has corank one, then $B = D A$ has corank one and the conclusion gives a nonsingular $(n - 1) \times (n - 1)$ principal submatrix.

Facts & Assumptions

Given: A real symmetric matrix $B$ of rank $r$ ; in the second assertion, also $A, D$ with $D$ positive diagonal and $B = D A$ .

[F1]

A symmetrizable generalized Cartan matrix has a positive diagonal symmetrizer $D$ for which $B = D A$ is symmetric (Symmetrizable generalized cartan matrix).

[F2]

For positive size, determinant is defined by the Leibniz formula; we additionally use the local empty-matrix convention stated above (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

[F3]

A real square matrix is invertible exactly when its determinant is nonzero (A finite square real matrix is invertible if and only if its determinant is nonzero); nonsingular means invertible (Invertible square matrices and similarity over a commutative ring).

[F4]

If a symmetric block matrix has an invertible leading block $C$ , its determinant is $\det (C)$ times the determinant of the Schur complement (For symmetric $M = (\begin{array}{cc} A & B \\ B^{T} & C \end{array})$ with $A$ invertible, a block-unitriangular congruence gives $A \oplus (C - B^{T} A^{- 1} B)$ and factors $\det M$ ).

[F5]

Matrix rank is row rank (Row space, column space, nullspace, row rank, column rank and matrix rank).

[F6]

A nonzero $k$ -rowed minor of a real matrix forces its rank to be at least $k$ (A matrix has rank at least $r$ exactly when it has a nonzero $r$ -rowed minor).

[F7]

Determinants are multiplicative, and the determinant of a diagonal matrix is the product of its diagonal entries (For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$ , The determinant of a triangular matrix is the product of its diagonal entries).

[F8]

Transpose reverses matrix products, and an invertible matrix has a unique inverse (Matrix arithmetic over a commutative ring is associative, unital and distributive, and transpose reverses products, Invertible square matrices and similarity over a commutative ring).

Proof

technique · Choose a maximal nonsingular principal block and use its Schur complement

1.1givenF2algebra

If $r = 0$ , then $B = 0$ and $J = \emptyset$ works by the stated empty-matrix convention. When $B = D A$ with positive diagonal $D$ , $A = 0$ as well, so the determinant identity is $1 = 1$ . Hence assume $r > 0$ .

1.2givenF2F3choosealgebra

Since $B \neq 0$ , either some $b_{i i} \neq 0$ , giving a nonsingular one-by-one principal submatrix, or all diagonal entries vanish and some $b_{i j} \neq 0$ with $i \neq j$ , giving the nonsingular two-by-two principal submatrix with determinant $- b_{i j}^{2}$ . Choose, from the finite family of principal submatrices with nonzero determinant, a set $J$ maximal under inclusion. Then $B_{J}$ is invertible by [F3]; put $k = ∣ J ∣$ .

2.1step 1.2F4algebra

For any $p \notin J$ , maximality makes $\det B_{J \cup {p}} = 0$ . Write $u_{p} = (b_{i p})_{i \in J}$ . The Schur-complement formula [F4] gives $0 = \det (B_{J}) (b_{p p} - u_{p}^{T} B_{J}^{- 1} u_{p})$ , so $b_{p p} - u_{p}^{T} B_{J}^{- 1} u_{p} = 0$ .

3.1step 2.1F4F8algebra

For distinct $p, q \notin J$ , maximality also gives $\det B_{J \cup {p, q}} = 0$ . Since $B_{J}^{T} = B_{J}$ , transposing $B_{J} B_{J}^{- 1} = I = B_{J}^{- 1} B_{J}$ and using [F8] shows that $(B_{J}^{- 1})^{T}$ is also a two-sided inverse of $B_{J}$ , hence $(B_{J}^{- 1})^{T} = B_{J}^{- 1}$ . The two-by-two Schur complement therefore has zero diagonal by step 2.1 and equal off-diagonal entries $t = b_{p q} - u_{p}^{T} B_{J}^{- 1} u_{q}$ . Its determinant is $- t^{2}$ , so [F4] and the fact that $\det B_{J} \neq 0$ imply $t = 0$ . Therefore every entry of the complementary block equals the corresponding entry of $W^{T} B_{J}^{- 1} W$ , where $W = B_{J, J^{c}}$ .

4.1step 3.1F5F6F8algebra

Partitioning by $J$ and $J^{c}$ , the equality in step 3.1 yields $B = (\begin{array}{c} I \\ W^{T} B_{J}^{- 1} \end{array}) B_{J} (\begin{array}{cc} I & B_{J}^{- 1} W \end{array})$ . By matrix multiplication, every row of $B$ is a linear combination of the $k$ rows of the right factor, so [F5] gives $r \leq k$ . Since $\det B_{J} \neq 0$ , $B$ has a nonzero $k$ -rowed minor, so [F6] gives $r \geq k$ . Hence $k = r$ .

5.1F1F5F7step 1.1step 4.1algebra∎

If $B = D A$ , diagonality gives $B_{J} = D_{J} A_{J}$ , and [F7] gives $\det B_{J} = \det D_{J} \det A_{J} = (\prod_{i \in J} d_{i}) \det A_{J}$ . The product is nonzero, so the already nonzero $\det B_{J}$ forces $\det A_{J} \neq 0$ . For a symmetrizable $A$ of corank one, positive diagonal row-scaling preserves rank, hence $r = n - 1$ .

Remarks

Symmetry is essential: $(\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array})$ has rank $1$ but no nonsingular one-by-one principal submatrix. Berkeley's Gabber–Kac example in Ch. 10 §10.4.2.6 assumes the positive-semidefinite corank-one case; the principal-minor argument above needs only symmetry and therefore also applies to indefinite symmetrizable matrices.

Depends on

Used by

Dependency tree · two levels

44 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Richard Borcherds, Mark Haiman, Theo Johnson-Freyd, Nicolai Reshetikhin and Vera Serganova, Berkeley Lectures on Lie Groups and Quantum Groups (book-length lecture notes, last updated 18 January 2024) (standard reference, not scraped)
Benjamin Enriquez, PBW and Duality Theorems for Quantum Groups and Quantum Current Algebras, Journal of Lie Theory 13 (2003), 21–64 (standard reference, not scraped)